study guide: hl books/group 4 - sciences... · 2020-05-12 · – fundamentals of organic chemistry...
TRANSCRIPT
S T U D Y G U I D E :
www.ib.academy
HL
IB Academy Chemistry Study GuideAvailable on learn.ib.academy
Author: Tim van Puffelen
Design Typesetting
This work may be shared digitally and in printed form,but it may not be changed and then redistributed in any form.
Copyright © 2020, IB AcademyVersion: CheHL.3.0.200227
This work is published under the Creative CommonsBY-NC-ND 4.0 International License. To view a copy of thislicense, visit creativecommons.org/licenses/by-nc-nd/4.0
This work may not used for commercial purposes other than by IB Academy, orparties directly licenced by IB Academy. If you acquired this guide by paying forit, or if you have received this guide as part of a paid service or product, directlyor indirectly, we kindly ask that you contact us immediately.
Laan van Puntenburg 2a3511ER, UtrechtThe Netherlands
[email protected]+31 (0) 30 4300 430
0
Welcome to the IB.Academy Study Guide for IB Chemistry High Level.
We are proud to present our study guides and hope that you will find them helpful. Theyare the result of a collaborative undertaking between our tutors, students and teachersfrom schools across the globe. Our mission is to create the most simple yetcomprehensive guides accessible to IB students and teachers worldwide. We are firmbelievers in the open education movement, which advocates for transparency andaccessibility of academic material. As a result, we embarked on this journey to createthese study guides that will be continuously reviewed and improved. Should you haveany comments, feel free to contact us.
For this Chemistry HL guide, we incorporated everything you need to know for yourfinal exam. The guide is broken down into manageable chapters based on the syllabustopics. The chapter is then further divided into subtopics.
For more information and details on our revision courses, be sure to visit our websiteat ib.academy. We hope that you will enjoy our guides and best of luck with your studies.
IB.Academy Team
3
TABLE OF CONTENTS
1. Quantitative chemistry 7– Types and states of matter – Chemical reactions – Moleconcept and chemical calculations
2. Atomic structure 21– Types of particles – Notation – Isotopes: abundance and Ar– Atomic shells/subshells/orbitals – Electromagneticspectrum – Ionization energies
3. Periodicity 31– The Periodic Table (PT) – Periodic trends – Transitionelements
4. Bonding 39– Metallic Bonding – Ionic bonding – Covalent bonding– Intermolecular forces – Properties of molecular compounds– Molecular orbitals – Hybridization: mixing atomic orbitals– Ozone and oxygen
5. Energetics 61– Temperature vs heat vs enthalpy – Energy diagrams– Hess’s law – Energy calculations – Energy cycles– Entropy
6. Kinetics 79– Collision Theory – Rate equation and reaction order
5
TABLE OF CONTENTS
7. Equilibrium 87– Dynamic equilibrium – Equilibrium law expression– States of matter – Le Chatelier’s principle – Equilibriumcalculations – Relation between∆G and Kc
8. Acids and bases 95– Acid and base definitions – Strong vs weak – pH scale– Buffers – pH curves – Acid Deposition
9. Redox 105– Oxidation states – Reactions – Reactivity– Electrochemical cells – The Winkler method and the BOD
10. Organic chemistry 115– Fundamentals of organic chemistry – Isomers – Reactions– Reaction mechanisms – Reactions overview andretrosynthesis
11. Measurement and data
processing
137
– Graphical Techniques – Spectroscopic identification
6
1QUANTITATIVE CHEMISTRY
1.1 Types and states of matter
all matter
mixtures substances
elements
non-metals (semimetals) metals
compounds
ioniccompounds
molecularcompounds
Classification of matter
Substance has a definite chemical composition and characteristic properties
Mixture contains multiple substances that retain their individual propertiesbecause they are not chemically bonded, which can be separated usingthe difference between the individual properties of each substance.
Homogeneous mixture: Heterogeneous mixture:Components are in the same phase,particles are distributed equally overthe mixture
Components are not all in the samephase, there are physical boundariesbetween the components.
e.g. solution, alloy e.g. suspension, emulsion
Classification of matter: type of element(s):metal metalionic compound metal + non-metalmolecular compound non-metal
Elements are atoms that have the same number of protons. Elementalsubstances contain one type of element (e.g., Na, Fe, H2, Cl2, S8, . . . )
Compounds at least two different elements combine to form a compound
The chemicalcomposition of asubstance is expressedin a chemical formula,which shows thenumber of each atom ina substance (e.g. H2O),or the ratio of ions in anionic compound (e.g.MgCl2).
7
QUANTITATIVE CHEMISTRY Types and states of matter
Note that for ioniccompounds only theempirical formula isused, because ionsassemble in a wholenumber ratio in alattice, but not asmolecules.
molecular formula structural formula empirical formula
example C2H4 C
H
H
C
H
H
CH2
showing number of atoms bonding between theatoms
simplest numberratio of atoms
Which compound has the empirical formula with the greatest mass?
molecular divisible empiricalformula by formula
A. C2H6 2 CH3
B. C4H10 2 C2H5 ← greatest mass
C. C5H10 5 CH2
D. C6H6 6 CH
Exam
ple.
Find the empirical formula given weight percentage composition
A compound is found to contain 64.80 % C, 13.62 % H, and 21.58 % O2 by weight. What
is the empirical formula for this compound?
1. Tabulate and assume 100 g C H O
grams 64.80g 13.62g 21.58g
2. Convert the masses to
moles (divide by the
atomic mass)
C H O
grams 64.80 g 13.62 g 21.58 g
moles 5.396mol 13.49mol 1.344mol
3. Divide by the lowest,
seeking the smallest
whole-number ratio
C H O
grams 64.80 g 13.62 g 21.58 g
moles 5.396 mol 13.49 mol 1.344 mol
simplest
ratio
4 10 1
4. Write the empirical formula C4H10O
8
QUANTITATIVE CHEMISTRY Types and states of matter 1
Derive the molecular formula from the empirical formula
From the previous, derive the molecular formula if the molecular mass is 222.4 g mol−1.
To determine themolecular formulainstead of the empiricalformula, the molecularmass must also begiven.1. molecular mass
mass of the empirical formula
222.4gmol−1
4 · 12.01+ 10 · 1.01+ 16.00gmol−1= 3
2. write the molecular formula C12H30O3 (since the molecule is 3 times
the mass of the empirical formula)
Phase changes and states of matter
liquid(l)
solid(s)
gas(g)
melting
evap
orat
ion deposition
freezing
sublimation
cond
ensa
tion
The state of a substance isindicated as (s), (l) or (g).
Additionally, when asubstance is dissolved inwater we can indicate thephase as (aq).
solid (s) liquid (l) gas (g)
movement of particles vibrational free movement free movement
distance between particles close close far apart
fixed volume yes yes no, compressable
fixed shape yes, rigid no no
9
QUANTITATIVE CHEMISTRY Chemical reactions
1.2 Chemical reactions
Chemical reaction a process that leads to the transformation of one set ofchemical substances to another, thus changing their chemical formulae
Combustion reaction a chemical reaction between a fuel and O2; whichcompounds form depends on which elements the fuel contains
fuel contains combustion product effect (environmental)
C CO2 (complete) greenhouse gas
CO (incomplete) toxic to animals
H H2O —
S SO2 acid rain (see chapter 8)
N NOx acid rain (see chapter 8)
Balancing and stoichiometry
To balance reactions we use the conservation of mass, which states that the number ofatoms before and after a reaction must be equal, and the conservation of charge, whichstates that the charge before and after a reaction must also be equal.
Stoichiometric coefficients the numbers placed in front of substances inorder to balance chemical reactions
Stoichiometry the quantitative relationships between substances in achemical reaction (molar ratios)
Balance the reaction: . . . C10H22 + . . . O2 −−→ . . . H2O+ . . . CO2
The trick to balancing chemical reactions is to balance elements in order of occurance.Both C and H occur in one substance before and after the reaction arrow, so balancethese first: 1 C10H22+ O2 −−→ 11 H2O+ 10 CO2
Next, balance O: 1 C10H22+312
O2 −−→ 11 H2O+ 10 CO2 (multiply by 2)
2 C10H22+ 31 O2 −−→ 22 H2O+ 20 CO2
Exam
ple.
10
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1
1.3 Mole concept and chemical calculations
Mole (n) the amount of substance which contains NA = 6.02× 1023 particles(atoms, molecules, ions, electrons, or other physical particles)
Avogadro’s constant (NA) 6.02× 1023 particles/mol
When buying eggs, you can request one dozen eggs from old-lady Mme. Oeuf.It is a convenient expression, since the packaging contains 12 eggs. So a dozenis an alternative way to express “12”.
The mole is analogously an alternative way to express the number of entities(6.02× 1023). This number is convenient to represent chemical amounts.
A dozen is a grouping of 12, so:2 dozen is a grouping of 24.
A mole is a grouping of 6.0× 1023, so:2 mole is a grouping of 1.2× 1024.
The mole concept is a necessity in chemical calculations. Since we constantly deal withhuge numbers of particles in chemistry, expressing the number of particles in moles ismore convenient. But more importantly, particles react and form in a particularstoichiometric ratio (molar ratio) in chemical reactions.
Take 2H2+O2 −−→ 2H2O; two molecules of H2 will react with one molecule of O2.
This does not mean that two grams of H2 will react with one gram of O2, since themasses of H2 and O2 molecules are not equal. This does mean that two moles of H2 willreact with one mole of O2, but also that 1.8× 10−3 mol H2 will react with0.9× 10−3 mol O2. Furthermore, in chemistry we constantly use huge ensembles ofmolecules. Using moles allows us to use much simpler numbers.
11
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations
Relative and average mass
Molecules have awell-defined number ofatoms, but ioniccompounds do notconsist of a particularnumber of ions, ratherof ions in a particularratio.
The relative masses are all measured relative to the atomic mass unit (u):defined as 1/12 of the atomic mass of a 12C atom. And they are all average: theweighed arithmetic mean of all isotopes and their terrestrial naturalabundancies.
Relative atomic mass (Ar ) the weighed mean of all the isotopes of oneelement and their natural abundances, relative to one atomic mass unit,which is 1/12 of the mass of a 12C atom
Relative molecular mass (Mr ) is the sum of the relative atomic masses of allthe atoms in the molecular formula
Relative formula mass (Mr ) applies to ionic compounds, and it is similar tothe relative molecular mass and also calculated in the same way.
Molar mass M the mass of a substance per one mole expressed in g mol−1
1.3.1 Unit conversion
Roadmap to chemical calculations
It is much easier to measure some physical quantity of a sample, such as its mass, than itis to count the number of particles in the sample. Therefore, you will have to convertvarious quantities to mole and back. The figure below gives an overview of the quantitiesthat can be converted to moles and back, and what other quantity you will need to do so.
molmol dm−3
concentration
no unit
N◦ of particles
g
mass
m3
volume (g)
Vsolvent Mr
NA
Vm
12
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1
Mole↔ number of particles
The number of particles can be expressed in the amount of moles, or in the number ofparticles. Because 1mol= 6.02× 1023 particles. The quantities express the same thing,but use different values in doing so. It’s like saying “a dozen eggs” to express 12 eggs.
The relationship between the number of particles and the amount in mol is given by:
N◦
of particles
NAmol−1
nmol
N◦ of particles= n ·NA
N◦ of particles=amount [no units]n =chemical amount [mol]NA =Avogadro’s constant 6.02× 1023 mol−1
How many N-atoms are there in 1.0× 10−2 mol NH3NO3?N◦ of particles= n ·NA= 1.0× 10−2 mol ·6.02× 1023 mol−1 = 6.02× 1021 molecules,per molecule there are 2 N-atoms (mole ratio molecules : N-atoms = 1 : 2),so 2 · 6.02× 1021 = 1.2× 1022 N-atoms.
What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?N◦ of particles= n ·NA= 0.20mol · 6.02× 1023 mol−1 = 1.2× 1023 molecules,per molecule there are 10 atoms (mole ratio molecules : atoms = 1 : 10),so 10 · 1.2× 1023 = 1.2× 1024 atoms.
Exam
ple.
Mole↔ gram
The molar mass (M ) can be calculated from the formula of the substance, which is the massof a substance per one mole particles (in g mol−1). It allows us to convert between themass and the amount of particles in moles.
The relationship between the amount in mol and mass in g is given by:
mg
nmol
Mg mol−1
m = n ·M
m=mass [g]n =chemical amount [mol]M =molar mass [g mol−1]
What is the amount in moles of 4.00 g in NaOH?
M (NaOH) = 22.99+ 16.00+ 1.01= 40.0gmol−1
n =mM=
4.00g
40.0gmol−1 = 0.100mol NaOH
Exam
ple.
13
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations
Mole↔ [concentration]
Solution a homogeneous mixture of a substance (the solute) dissolved inanother substance (the solvent)
(Molar) concentration (C) is the amount of solute (in mol) per unit volume(in dm3), often written using [. . . ], and expressed in mol dm−3
Standard solution a solution with a known concentration of solute
The solute concentration isindependent of the solvent volume.When we dissolve 4 mol sugar in 2 dm3
water, it has a particular ‘sweetness’.This sweetness is a measure of theconcentration of sugar in water. If wewould poor out 1 dm3 from thesolution, it would still be equally sweet(same concentration) but only containhalf the sugar content (half the amountin moles).
2 dm3
4mol2dm3 = 2moldm−3
1 dm3
2mol1dm3 = 2moldm−3
The relationship between the amount in mol and concentration in mol dm−3 is given by:
nmol
Cmol dm−3
Vsolventdm3
n =Vsolvent ·C
n =chemical amount [mol]Vsolvent= solvent volume [dm3]C =concentration [mol dm−3]
It’s very useful toremember thatcm3 ·moldm−3 =mmol
What amount of NaCl (in moles) is required to prepare 250 cm3 of a0.200 mol dm−3 solution?n =Vsolvent ·C = 250cm3 · 0.200moldm−3 = 50mmol
Which solution contains the greatest amount (in mol) of solute?Vsolvent ·C = n in mol
A. 10.0 cm3 of 0.500 mol dm−3 NaCl 10.0 ·0.500 = 5mmolB. 20.0 cm3 of 0.400 mol dm−3 NaCl 20.0 ·0.400 = 8mmolC. 30.0 cm3 of 0.300 mol dm−3 NaCl 30.0 ·0.300 = 9mmol ←D. 40.0 cm3 of 0.200 mol dm−3 NaCl 40.0 ·0.200 = 8mmol
Exam
ple.
14
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1
Mole↔ volume gas
Avogadro’s law equal volumes of all gases, at the same temperature andpressure, have the same number of molecules
Assuming the gas is anideal gas, Avogadro’slaw is tested quiteoften: it should beunderstood that themolar ratio can beapplied to gas volumes.
Molar volume (Vm) the volume of one mole gas, expressed in dm3 mol−1 orm3 mol−1, at a particular pressure and temperature.
The volume of an ideal gas at constant temperature and pressure is proportional to thenumber of particles (in moles). So when twice the number of particles are placed into acylinder, then the volume becomes twice as large. The volume of one mole gas (molarvolume) at STP is 22.7 dm3 mol−1.
22.7 dm 45.4 dm
273 K
100 kPa
The relationship between the amount in mol and gaseous volume in dm3 is given by:
Vgasdm3
nmol
Vmdm3 mol−1
Vgas = n ·Vm
n =chemical amount [mol]Vgas=gas volume [dm3]Vm =molar volume [dm3 mol−1]
Calculate the volume of nitrogen gas produced by the decomposition of2.50 mol of NaN3(s) at STP in the reaction 2NaN3(s) −−→ 2 Na(s)+ 3 N2(g).
Since the molar ratio NaN3 : N2 = 2 : 3, 2.50mol2 × 3= 3.75mol N2 forms.
At STP Vm = 22.7dm3 mol−1, Vgas = n ·Vm = 3.75 · 22.7= 85.1dm3
Exam
ple.
15
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations
1.3.2 Gas laws and ideal gases
An ideal gas is a theoretical gas that assumes that: the volume of particles is negligiblecompared to the volume of the surrounding empty space, and no kinetic energy is lost inthe collisions between the particles. Whether these assumptions are justified is outside thescope of the IB syllabus, so from now on we will treat all gases as ideal gases.
Pressure the force exerted by the collisions of particles on the walls of itscontainer
pressure at sea level = 100kPa= 1.00× 105 Pa
Temperature the average kinetic energy of particles
Tin K = Tin ◦C+ 273 and Tin ◦C = Tin K− 273
STP standard temperature and pressure: 273K and 100kPa
SATP standard ambient temperature and pressure: 298K and 100kPa
The ideal gas law assumes ideal gas behaviour, and it is an equation that relates thepressure, volume, amount in moles and the temperature of a gas. Critically, SI units mustbe used in the ideal gas law:Memorize the ideal gas
law: pV = nRT . Forpaper 2 you will have itin the databook, butyou will also need it forpaper 1 questions! pV = nRT
p =pressure [Pa]V =volume [m3]n =amount of substance [mol]R = ideal gas constant 8.31 J K−1 mol−1
T = temperature [K]
Using the ideal gas law, verify that Vm at STP is 22.7 dm3 mol−1.
Molar volume (Vm) = the volume (V ) per mole (n), or Vm =Vn
.
Rearrange the ideal gas law:Vn=
RTp=Vm
Vm =RT
p=
8.31 · 273K1.00× 105 Pa
= 2.27× 10−2 m3 mol−1 = 22.7dm3 mol−1
Exam
ple.
16
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1
Questions involving the ideal gas law in paper 1 are often presented ‘at constant mass’;this means that the number of moles n is kept constant. R is also a constant. Rearrangingthe ideal gas law with the variables to one side, and the constants to the other yields:pVT= nR.
Memorize this formulaor how to derive it,because it will allowyou to answer allquestions regarding gaslaws!
Since only p, V and T are allowed to change, nR will remain constant.
Mathematically, two situations (1) and (2) can be related by:
nR=p1V1
T1=
p2V2
T2
Fromp1V1
T1=
p2V2
T2we can derive Boyle’s law, Charles’ law and Gay-Lussac’s Law (and
Wikipedia knows which is which). In each of the three gas laws, one of the quantities ( p,V or T ) is fixed, as well n (‘at constant mass’).
constant pressure constant volume constant temperature
derivation ofgas law
ZZp1V1
T1=ZZ
p2V2
T2
V1
T1=
V2
T2
p1@@V1
T1=
p2@@V2
T2p1
T1=
p2
T2
p1V1
@@T1=
p2V2
@@T2
p1V1= p2V2
relation V ∝ T p∝ T p∝ 1V
graph
T
V
T
P
P
V
At constant temperature, sketch a diagram that shows how p changes when Vchanges.p1V1 = p2V2 holds, so put in some numbers to figure out the relation.Assume p1V1 = 1 · 1= 1:
V
P
P ×V = 1= constant
2× 12= 1= constant
1× 1= 1= constant12× 2= 1= constant
�
12
,2�
(1,1)�2,
12
�
Exam
ple.
17
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations
1.3.3 Chemical calculations
The basics of all chemical calculations can be summarized as follows:
Known substance X Known substance Y1 2 3
× molar ratio
Requiredmolar mass
dm3 solutionmolar volume
g
mol L−1
m3
N◦ of particles
mass
concentration
volume
g
mol L−1
m3
N◦ of particles
mass
concentration
volume
molsubstance
X
molsubstance
Y
Performing chemical calculations
5.0 g of CH4 undergoes complete combustion. Calculate the volume of the resulting
gases under STP assuming that water forms as a gas.
1. Note the reaction equation and list
the information given
CH4+ 2O2 −→ CO2+ 2H2O 5.0 g CH4
2. Convert units to mole CH4Mm = 12.0+ 4 · 1.01= 16.04gmol−1
5.016.04
= 0.312mol CH4
3. Use the molar ratio to convert to the
number of moles of the substance(s)
asked for
The volume of all the gases that form is
required. Per 1 mol CH4, 3 mol gas forms
0.312mol CH4 ·31= 0.935mol gas.
4. Convert moles to required units Under STP Vm = 24.5dm3 mol−1,
0.935mol · 24.5dm3 mol−1 = 22.9dm3
5. Check significant figures and units Looking back to step 1 the amount of CH4 is
given in two significant figures, so the answer
should also be written using two significant
figures −→ 23 dm3
18
QUANTITATIVE CHEMISTRY Mole concept and chemical calculations 1
Limiting and excess reactant
12 table legs and 4 table tops are stored in the warehouse of a table factory. Our intuitiontells us that we cannot make 4 tables, since it would require 16 table legs. So even thoughmore table legs are available than table tops, due to the ratio in which they are needed thetable legs are the limiting reactant and the table tops are in excess. Doing chemistry wedo not have the same intuition, but we perform the same math.
Theoretical yield the maximum quantity of product that can be obtainedfrom given quantities of reactants, assuming completion
Limiting reactant the reactant that determines the theoretical yield of aproduct, after the reaction is complete there will be none left
Excess reactant the reactant that is not used up by the reaction, after thereaction is complete this substance will still be present
All calculations must be done using the amount of the limiting reactant, since itdetermines how much product will be made (i.e. the available 12 table legs determinethat the theoretical yield of tables is 3).
Identify the limiting reactant
4.22 g Al reacts with 25.0 g Br2 in the following reaction: 2Al+ 3Br2 −→ 2AlBr3. Identify
the limiting reactant and use it to determine the theoretical yield of AlBr3.
1. Convert units to moles 4.22g
26.98gmol−1= 0.156mol Al
25.0g
2 · 79.90gmol−1= 0.156mol Br2
2. Divide the number of moles of each
reactant by its reaction coefficient. The
reactant with the lowest result will be the
limiting reactant.
Al:0.156mol
2= 0.078
Br2:0.156mol
3= 0.052
So Br2 is the limiting reactant.
3. Use the number of moles of the limiting
reactant from step 1 and the molar ratio
to calculate the number of moles of the
requested substance
The molar ratio of AlBr3 : Br2 = 2 : 3, so
0.156mol · 23= 0.104mol AlBr3
19
2ATOMIC STRUCTURE
2.1 Types of particles
Particles
Atoms
no chargeN◦p+ = N◦e−
protons
elements havethe same Z/p+
neutrons
isotopes have adifferent N◦ n0
electrons
core e– vsvalence e–
Ions
charged+/−: atomswith e– defi-
ciency/sufficiency
+ cation
p+ > e−
− anion
p+ < e−
Molecules
cluster ofcovalently
bonded atoms
Atoms contain subatomic particles: protons, neutrons and electrons
Nucleus + protons and neutrons form the atom’s nucleus
Electron cloud − electrons occupy the space outside the nucleus inshells/subshells/orbitals
notation mass relative mass charge relative charge
proton p+ 1.67× 10−24 g 1 u 1.60× 10−19 C +1neutron n0 1.67× 10−24 g 1 u 0 C 0electron e− 9.11× 10−28 g ≈0 u −1.60× 10−19 C −1
21
ATOMIC STRUCTURE Notation
Atom charge = 0. the smallest constituent unit of ordinary matter that hasthe properties of a chemical element
Ion charge 6= 0, so the number of e− 6= the number of p+
Negative ion/anion − contains more e− than p+
Positive ion/cation + contains less e− than p+
Element all atoms of the same element have the same number of p+
(i.e. the same atomic number Z)
Isotopes atoms of the same element but with a different number of n0,resulting in a different mass number AX.
anion: a negative ion
cation: is pawsitive
Isotopes have the samechemical properties,but different physicalproperties
2.2 Notation
C6
12
massnumber A =N◦p++N◦n0
charge: e− deficiency/sufficiency
chemical symbol, determined by:
atomic number Z =N◦p+
This atom has 8 p+ and 10 n0, what is the chemical notation for this ion?
nucZ=N◦p+ = 8A=N◦p++N◦n0 = 8+ 10= 18charge=N◦p+−N◦e− = 8− 10=−2
188O
2−
− ions have extra e−!
Find the symbol, Z, p+, n0, e− for: 115X, 19
9X– and 27Al3+
Symbol and Z N◦ p+ N◦ n0 N◦ e−
115X Z= 5, so B 5 11− 5= 6 5
199X
– Z= 9, so F 9 19− 9= 10 9+ 1= 1027Al3+ Al, so Z= 13 13 27− 13= 14 13− 3= 10
Exam
ple.
22
ATOMIC STRUCTURE Isotopes: abundance and Ar 2
2.3 Isotopes: abundance and Ar
Relative atomic mass (Ar ) the weighed mean of all the isotopes of oneelement and their natural abundances, relative to one atomic mass unit,which is 1/12 of the mass of a 12C atom
Ar =
fractionalabundanceof isotope 1
× mass ofisotope 1
+
fractionalabundanceof isotope 2
× mass ofisotope 2
+ . . .
1u= 1gmol−1
Calculate Ar of bromine, given that the abundancies of 79Br and 81Br are50.69 % and 49.31 %.Ar (Br) = 50.69% · 79Br+ 49.31% · 81Br
= 0.5069 · 79+ 0.4931 · 81
= 79.90gmol−1
Calculate the abundancies of 69Ga and 71Ga, given these are the only stableisotopes of Ga and Ar = 69.72 gmol−1
Since 69Ga and 71Ga are the only stable isotopes we can say that:fractional abundance 69Ga+ fractional abundance 69Ga= 1(or 100 %)If we let x = fractional abundance 69Ga then fractional abundance 71Ga= 1− x
Ar (Ga) = x · 69Ga+(1− x) · 71Ga
69.72gmol−1 = x · 69+(1− x) · 71= 69x + 71− 71xx = 0.64
So the abundance of 69Ga is 64 % and the abundance of 71Ga is 100%− 64%= 36%
Exam
ple.
23
ATOMIC STRUCTURE Isotopes: abundance and Ar
Mass spectrometer
A mass spectrometer is an analytical instrument that can measure the mass of eachisotope in a sample. So if a sample of lead, Pb, is injected into the device, the followingspectrum and relative abundances will result:
Mass/Charge (m/z)
Rel
ativ
eab
unda
nce
203 204 205 206 207 208 209
102030405060708090
100
3.9
46.2 42.3
100
RelativeIsotope abundance204Pb 3.9206Pb 46.2207Pb 42.3208Pb 100
The data of the mass spectrum allows us to calculate the average atomic mass byweighing the isotopic mass against its relative abundance. Note that the abundance is notgiven as a %, so we have to divide by the sum of all the relative abundances.
Ar (Pb)=3.9 · 204+ 46.2 · 206+ 42.3 · 207+ 100 · 208
3.9+ 46.2+ 42.3+ 100= 207.2gmol−1
24
ATOMIC STRUCTURE Atomic shells/subshells/orbitals 2
2.4 Atomic shells/subshells/orbitals
Electron shell n = 1, 2, 3 . . . principal energy level which each contains 2n2
electrons, further divided in a number of subshells
Subshells s, p, d, f each subshell has a particular number of orbitals, andeach has its own geometry
Atomic orbital region with a specific geometry that can host two electronsof opposite spin
shell max. N◦ of e− N◦ of orbitals electronn 2n2 s p d f total configuration
1 2 · 12 = 2 1 — — — 1 1sx
2 2 · 22 = 8 1 3 — — 4 2sx 2px
3 2 · 32 = 18 1 3 5 — 9 3sx 3px 3dx
4 2 · 42 = 32 1 3 5 7 16 4sx 4px 4dx 4fx
Electron shell is alsooften referred to as themain energy level
+
The principal electron shells can be imagined as anonion: the first shell (n = 1) is closest to the nucleus andeach of the following shells is further away. The shellsare numbered: n = 1, n = 2, . . .
Every shell contains the s subshell, from the secondshell (n = 2) forward all shells contain the p subshell,from the third shell (n = 3) all shells contain the dsubshell etc.
The shape of the subshells
Every s subshell consist of one spherical orbital, which is further away from the nucleusthe higher the shell number. The p subshell always contains 3 orbitals which are alignedalong the xy z-axis. Since the second shell (n = 2) contains the 2s 2p subshells which canhost 2 and 6 electrons respectively, the maximum total number of electrons in the secondshell is 8.
s orbital px orbital py orbital pz orbital
25
ATOMIC STRUCTURE Atomic shells/subshells/orbitals
2.4.1 Electron configuration
Electron configuration shows the number ofe– in each subshell in the groundstate(the lowest energy state)
Aufbau principle = 1s 2s 2p 3s 3p 4s 3d 4pelectrons are placed into orbitals withlowest energy first, which are notnecessarily orbitals closest to thenucleus. Importantly: electrons occupy4s before 3d orbitals.
Pauli’s exclusion principle per orbital amaximum of 2 e– with opposite spin areallowed.
Hund’s rule instead of forming pairs in thesame orbital, electrons rather occupyempty orbitals to minimize repulsion.
E
1s��
2s��
3s��
4s��
�� �� ��
2p
�� �� ��
3p
4p �� � � � �
3d
Write the electron configuration of atoms
Write the full and condensed electron configuration for Fe.
1. Determine the total number of e− 26Fe, is an atom so:
number of p+ = number of e – = 26
2. Allocate each electron to each subshell
according to the Aufbau principle. Place
max. 2 e – in the s subshells, max. 6 e – in
the p subshells, max. 10 e – in d subshell.
1s2 2s2 2p6 3s2 3p6 4s2 3d6
Abbreviated form:
[Ar] 4s2 3d6
Note that when askedto write the full electronconfiguration, thecondensed form isincorrect!
To write the electron configuration of ions, we have to add or remove electrons from theelectron configuration of the atom equal to the charge of the ion. As the 3d sublevelbecomes populated with electrons, the relative energies of the 4s and 3d fluctuate relativeto one another and the 4s ends up higher in energy as the 3d sublevel fills. This means that4 s electrons are removed prior to 3d electrons.
26
ATOMIC STRUCTURE Electromagnetic spectrum 2
Write the electron configuration of ions
Write the condensed electron configuration for Fe3+
1. Write the electron configuration of the
atom[Ar]4s23d6
2. Add electrons / remove electrons from
the outermost shell (4 s before 3d)
Fe3+ has 3 fewer e – than the atom. First
remove two 4s electrons, and then
remove one 3d electron: [Ar] 3d5
d4 and d9 exceptions
The situation when subshells are completely filled, orhalf-filled, is energetically favoured. An electron fromthe 4s subshell can be promoted to attain a half-filledd-subshell (d5) or full d-subshell (d10).
incorrect correctCr [Ar]��4s23d4 [Ar] 4s1 3d5
Cu [Ar]��4s23d9 [Ar] 4s1 3d10
Memorize these twoexceptions becausethey have been testedoften!
2.5 Electromagnetic spectrum
Electromagnetic radiation a form of energy that propagates through spaceat the speed of light as electromagnetic waves, or photons
Ephoton = hν =hcλ
Ephoton=energy of a photon [J]h =Planck’s constant 6.63× 10−34 J sν or f = frequency [s−1]λ =wavelength [m]
10−5 10−4 10−3 10−2 10−1 100 101 102 103 104 105 106 107 108 109
Wavelength λ (µm)
γ -rayX-rayUltra violetVisible lightInfraredMicro waveRadio wave
Visible light (Vis) is an example of electromagnetic (EM) radiation. The colour of light istied to the amount of energy of a photon. But visible light is only a small part of the EMspectrum; at the higher energy end of the spectrum we find ultraviolet (UV), X-ray andγ -ray, at the lower energy end of the spectrum we find infrared (IR), microwaves andradio waves.
27
ATOMIC STRUCTURE Electromagnetic spectrum
Electron energy levels
The energy level of an electron depends on which atomic orbital it occupies. The lowestenergy level is called the groundstate; an electron can move to a higher energy level(excited state) by absorption of a photon. And similarly, an electron can move from anexcited state to a lower energy level by emitting a photon.
The transition between electron energy levels is only possible when the electron absorbsor emits a photon with exactly the same amount of energy as the difference between theenergy levels. Energy transitions are discrete: of a particular amount of energy.
+
proton
e−photon emitted
lower energy state higher energy state
When excited electrons ‘fall’ from a higher to a lower energy state, photons with adiscrete amount of energy are emitted. The emission spectrum of atoms is a linespectrum: only light of a particular colour (discrete energy) is emitted.
Continuous spectrum
380 400 420 440 460 480 500 520 540 560 580 600 620 640 660 680 700 720 740 760 780
Hydrogen
380 400 420 440 460 480 500 520 540 560 580 600 620 640 660 680 700 720 740 760 780
28
ATOMIC STRUCTURE Ionization energies 2
Hydrogen energy levels
n = 1
n = 2
n = 3n = 4n = 5n = 6
VisUV
E
n = 1groundstate
n = 2
n = 3
n = 4
n = 5n = 6n =∞
ionised atom
visible lightn = · · · → n = 2
ultravioletn = · · · → n = 1
infraredn = · · · → n = 3
excitedstates
Electrons that ‘fall’ to the groundstate (n = 1) emit photons with the greatest amount ofenergy (UV radiation). The length of the arrows is proportional to the amount of energy.Electrons that ‘fall’ to n = 2 emit visible light and to n = 3 emit infrared radiation.
Also note that the energy levels converge at higher energy: the difference between theenergy levels becomes smaller up to the point where the difference is 0. The energydifference between the more energetic photons is increasingly smaller. Therefore linespectra converge at higher energy.
2.6 Ionization energies
First ionization energy the energy required to remove one mole electronsfrom one mole of gaseous atoms, to produce one mole of gaseous 1+ ions.
Evidence for shells
From He over Ne to Ar: downthe group it becomes easier toremove an electron since it isfurther away from the nucleus(in a higher shell).
From He to Li: the first electron ina new shell is easily removedbecause the effective nuclearcharge is low (+1), so it isweakly attracted.
atomic number
first
ioni
zatio
nen
ergy
(kJm
ol−
1 )
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
H
He
Li
Be
BC
N
O
F
Ne
Na
Mg
AlSi
PS
ClAr
K
Ca
29
ATOMIC STRUCTURE Ionization energies
From Li to Ne: across a period it becomes harder to remove an electron since theeffective nuclear charge increases, so the attraction increases.
Evidence for subshells
From Be [He]s2 to B [He]s2p1: an electron is added to the p-subshell, which has aslightly higher energy than the s-subshell, meaning it can be removed more easily.
From N [He]s2p3 to O [He]s2p4: an electron is paired in a p-orbital causing increasedrepulsion, meaning it can be removed more easily.
Successive ionisation energies
number of electrons removed
log(IE)
1 2 3 4 5 6 7 8 9
Note that y-axis isplotted logarithmically,so abrupt changes inthe y direction arecompressed but are stilleasy to identify.
In the figure above we see the abrupt change occurring after the 3rd electron, whichmeans that it is much harder to remove electron 4 than electron 3. All elements ingroup 13 (B, Al,Ga. . . ) will have this characteristic, since after removing 3 electrons,removing the 4th will break the noble gas configuration.
In the figure above we see the abrubt change occurring after the 3rd electron, whichmeans that it is much harder to remove electron 4 than electron 3. All elements ingroup 13 (B, Al,Ga. . . ) will have this characteristic, since after removing 3 electrons,removing the 4th will break the noble gas configuration.
30
3PERIODICITY
3.1 The Periodic Table (PT)
Period is a row in the periodic table, and it represents the principal electronshells (n = 1,2, . . . )
Group is a column in the periodic table, and it groups elements with similarchemical properties due to having the same number of valance e–
Valence electrons arejust the outermostelectrons
H1 1.0079
Hydrogen
Li3 6.941
Lithium
Na11 22.990
Sodium
K19 39.098
Potassium
Rb37 85.468
Rubidium
Cs55 132.91
Caesium
Fr87 223
Francium
Be4 9.0122
Beryllium
Mg12 24.305
Magnesium
Ca20 40.078
Calcium
Sr38 87.62
Strontium
Ba56 137.33
Barium
Ra88 226
Radium
Sc21 44.956
Scandium
Y39 88.906
Yttrium
La57 138.91
Lanthanide
Ac89 227
Actinide
Ti22 47.867
Titanium
Zr40 91.224
Zirconium
Hf72 178.49
Halfnium
V23 50.942
Vanadium
Nb41 92.906
Niobium
Ta73 180.95
Tantalum
Cr24 51.996
Chromium
Mo42 95.94
Molybdenum
W74 183.84
Tungsten
Mn25 54.938
Manganese
Tc43 96
Technetium
Re75 186.21
Rhenium
Fe26 55.845
Iron
Ru44 101.07
Ruthenium
Os76 190.23
Osmium
Co27 58.933
Cobalt
Rh45 102.91
Rhodium
Ir77 192.22
Iridium
Ni28 58.693
Nickel
Pd46 106.42
Palladium
Pt78 195.08
Platinum
Cu29 63.546
Copper
Ag47 107.87
Silver
Au79 196.97
Gold
Zn30 65.39
Zinc
Cd48 112.41
Cadmium
Hg80 200.59
Mercury
Rf104 261
Rutherfordium
Db105 262
Dubnium
Sg106 266
Seaborgium
Bh107 264
Bohrium
Hs108 277
Hassium
Mt109 268
Meitnerium
Ds110 281
Darmstadtium
Rg111 280
Roentgenium
Uub112 285
Ununbium
B5 10.811
Boron
Al13 26.982
Aluminium
Ga31 69.723
Gallium
In49 114.82
Indium
Tl81 204.38
Thallium
C6 12.011
Carbon
Si14 28.086
Silicon
Ge32 72.64
Germanium
Sn50 118.71
Tin
Pb82 207.2
Lead
N7 14.007
Nitrogen
P15 30.974
Phosphorus
As33 74.922
Arsenic
Sb51 121.76
Antimony
Bi83 208.98
Bismuth
O8 15.999
Oxygen
S16 32.065
Sulphur
Se34 78.96
Selenium
Te52 127.6
Tellurium
Po84 209
Polonium
F9 18.998
Flourine
Cl17 35.453
Chlorine
Br35 79.904
Bromine
I53 126.9
Iodine
At85 210
Astatine
He2 4.0025
Helium
Ne10 20.180
Neon
Ar18 39.948
Argon
Kr36 83.8
Krypton
Xe54 131.29
Xenon
Rn86 222
Radon
Uut113 284
Ununtrium
Uuq114 289
Ununquadium
Uup115 288
Ununpentium
Uuh116 293
Ununhexium
Uus117 292
Ununseptium
Uuo118 294
Ununoctium
La57–71
Lanthanum
Ce58 140.12
Cerium
Pr59 140.91
Praseodymium
Nd60 144.24
Neodymium
Pm61 145
Promethium
Sm62 150.36
Samarium
Eu63 151.96
Europium
Gd64 157.25
Gadolinium
Tb65 158.93
Terbium
Dy66 162.50
Dysprosium
Ho67 164.93
Holmium
Er68 167.26
Erbium
Tm69 168.93
Thulium
Yb70 173.04
Ytterbium
Lu71 174.97
Lutetium
Ac89–103
Actinium
Th90 232.04
Thorium
Pa91 231.04
Protactinium
U92 238.03
Uranium
Np93 237
Neptunium
Pu94 244
Plutonium
Am95 243
Americium
Cm96 247
Curium
Bk97 247
Berkelium
Cf98 251
Californium
Es99 252
Einsteinium
Fm100 257
Fermium
Md101 258
Mendelevium
No102 259
Nobelium
Lr103 262
Lawrencium
XZ mass
Name
1
2
3
4
5
6
7
1 IA
2 IIA
3 IIIA 4 IVB 5 VB 6 VIB 7 VIIB 8 VIIIB 9 VIIIB 10 VIIIB 11 IB 12 IIB
13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA
18 VIIIA
Alkali MetalTransition MetalHalogenNoble GasLanthanideActinide
Element type
Metals the majority of elements (in the figure: from blue to green), found on the left in the PT.Non-metals form mostly molecules, (in the figure: from orange to red), found on the right in the PT.Metalloids have intermediate properties
The division between elements that are metals and non-metals starts between Al (whichis a metal) and B, and staircases down to the right. Later on it will be assumed knowledgeand it is crucial to quickly tell if an element is metallic or not. If you have never heard
of the element, chancesare that it’s a metal.
31
PERIODICITY Periodic trends
Subshells & blocks
The elements in the periodic table can be divided into four blocks, based on theirelectronic configuration. Since elements down a group have the same number of valenceelectrons, they will also have the same outermost subshell configuration.
Take the alkali metals for example: each has a single electron in the outermost shell, buteach in a shell further away. The electron configurations of Li, Na and K are [He] 2s1,[Ne] 3s1 and [Ar] 4s1 respectively. The shell number that contains those electrons can beread off from the period number.
Note that H has the s1
configuration but is nota metal, and He has s2
configuration so wecould consider placingit in above Be. Butsince the valence shellof He is completely fullits propertiescorrespond much closerto the noble gases.
s-block d-block p-block
HHydrogen
LiLithium
NaSodium
KPotassium
RbRubidium
CsCaesium
FrFrancium
BeBeryllium
MgMagnesium
CaCalcium
SrStrontium
BaBarium
RaRadium
1
2
3
4
5
6
7
1 IA
2 IIA
s1 s2
ScScandium
YYttrium
LaLanthanide
AcActinide
TiTitanium
ZrZirconium
HfHalfnium
VVanadium
NbNiobium
TaTantalum
CrChromium
MoMolybdenum
WTungsten
MnManganese
TcTechnetium
ReRhenium
FeIron
RuRuthenium
OsOsmium
CoCobalt
RhRhodium
IrIridium
NiNickel
PdPalladium
PtPlatinum
CuCopper
AgSilver
AuGold
ZnZinc
CdCadmium
HgMercury
4
5
6
7
3 IIIA 4 IVB 5 VB 6 VIB 7 VIIB 8 VIIIB 9 VIIIB 10 VIIIB 11 IB 12 IIB
d1 d2 d3 d4 d5 d6 d7 d8 d9 d10
BBoron
AlAluminium
GaGallium
InIndium
TlThallium
CCarbon
SiSilicon
GeGermanium
SnTin
PbLead
NNitrogen
PPhosphorus
AsArsenic
SbAntimony
BiBismuth
OOxygen
SSulphur
SeSelenium
TeTellurium
PoPolonium
FFlourine
ClChlorine
BrBromine
IIodine
AtAstatine
HeHelium
NeNeon
ArArgon
KrKrypton
XeXenon
RnRadon
1
2
3
4
5
6
13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA
18 VIIIA
p1 p2 p3 p4 p5 p6
f-block
LaLanthanum
CeCerium
PrPraseodymium
NdNeodymium
PmPromethium
SmSamarium
EuEuropium
GdGadolinium
TbTerbium
DyDysprosium
HoHolmium
ErErbium
TmThulium
YbYtterbium
LuLutetium
AcActinium
ThThorium
PaProtactinium
UUranium
NpNeptunium
PuPlutonium
AmAmericium
CmCurium
BkBerkelium
CfCalifornium
EsEinsteinium
FmFermium
MdMendelevium
NoNobelium
LrLawrencium
The elements in the first row are called lanthanides, and in the second row actinides.
3.2 Periodic trends
The nucleus of fluorinehas a charge of 9+, the1st shell is full, reducingthe effective attractivepower of the nucleus tothe valence electrons to7+.
Periodic trends are governed by the balance between attractive and repulsive electrostaticforces between the nucleus and the valence electrons.
Shielding electrons in lower full electronshells reduce the + -charge that thevalence electrons experience
Effective nuclear charge (Zeff) the net+ charge that valence electrons
experience.
non-shieldingouter shell (7e−)
neon shell
helium filled shell(2e−, shielding)
nucleus (9+)
Zeff = Z −N◦shielding electrons Zeff = 9− 2=+7
32
PERIODICITY Periodic trends 3
Determine the effective nuclear charge (Zeff) for O, F, and Cl
8O 1s2 2s2 2p4, it has 2 e− in the filled 1st shell (shielding).
Zeff = Z −N◦ shielding electrons = 8− 2=+6
Notice how the number
of shielding e− doesnot change across theperiod!
9F 1s22s22p5, so 2 e− in the filled 1st shell (shielding).
Zeff = Z −N◦ shielding electrons = 9− 2=+7Notice how Zeff doesnot change down agroup!17Cl 1s22s22p63s23p5, so 10 shielding electrons (2+ 8e− in the 1st and 2nd shells).
Zeff = Z −N◦ shielding electrons = 17− 10=+7
Exam
ple.
The valence electrons experience an attractive force to the nucleus proportional to:
The effective nuclear charge (Zeff) The higher the effective nuclear charge, thestronger the valence electrons are attracted to the nucleus. The effective nuclearcharge increases→ a period.
The distance The further away the valence electron is from the nucleus, the weaker it isattracted. The distace between the valence shell and the nucleus increases withincreasing shell number, so ↓ a group.
The valence electrons mutually repel each other. This repelling force increases whenthere are more electrons in the valence shell.
Comparatively, the effect of the attractive forces is stronger than the repelling forces,which means that only when the effective nuclear charge and the shell stays the same dowe use arguments based on the repelling forces between valence electrons (ions).
To summarize, attraction between the + nucleus and the − valence electrons increases:→ the period Zeff increases, causing the valence electrons to experience stronger
attraction to the nucleus. The valence shell number is the same, so the electronsare at the same (approximate) distance.
↑ the group e− are closer to the nucleus, causing the valence electrons to experiencestronger attraction to the nucleus. The Zeff stays the same in the same group.
H
Li
Na
K
Rb
Cs
Fr
Be
Mg
Ca
Sr
Ba
Ra
Sc
Y
La
Ac
Ti
Zr
Hf
V
Nb
Ta
Cr
Mo
W
Mn
Tc
Re
Fe
Ru
Os
Co
Rh
Ir
Ni
Pd
Pt
Cu
Ag
Au
Zn
Cd
Hg
Rf Db Sg Bh Hs Mt Ds Rg
B
Al
Ga
In
Tl
C
Si
Ge
Sn
Pb
N
P
As
Sb
Bi
O
S
Se
Te
Po
F
Cl
Br
I
At
He
Ne
Ar
Kr
Xe
Rn
33
PERIODICITY Periodic trends
Attraction ↑ (between the nucleus and the valence e– )
→ a period, because Zeff increases (larger charge difference)↑ a group, because distance decreases (larger distance difference)
Atomic radius the distance from the nucleus to the valence electron(s).
– When attraction ↑, the atomic radius ↓.
Ionic radius the distance from the nucleus to the valence electron(s).
Zeff and the shell number stay the same, so the attraction stays the same. Onlythe repelling forces between the valence e– changes:
– In + ions a number e− are removed. All things equal, the mutuallyrepelling forces between valence electrons decreases. The larger the+ charge, the smaller the ion.
– In − ions a number of e− are added. All things equal, the mutuallyrepelling forces between valence electrons increases. The larger the+ charge, the larger the ion.
Electronegativity a measure of the ability of an atom to attract e− in acovalent bond, i.e. loosely how greedy an element is for electrons
– When attraction ↑, the electronegativity ↑.
First ionization energy the energy change when one mole electrons areremoved from one mole gaseous atoms, forming one mole gaseous 1+ions
– When attraction ↑, the first ionization energy ↑.
First electron affinity the energy change when one mole of gaseous atomsacquire one mole electrons, forming one mole gaseous 1- ions.
– When attraction ↑, the first electron affinity ↑.
The effect of adding orremoving one electronwhile the attractionremains equal issignificant. It’s safe toassume that + ions arealways smaller thanatoms, and - ionsalways larger.
The energies of the firstionisation energy andelectron affinity bothincrease when theattraction increases: iteither requires moreenergy to remove theelectron or releasesmore when it’sacquired.
Read: Na+ is smallerthan Ne is smaller thanF – , or the reverse
Order Ne, F– and Na+ in decreasing size. Explain why these are isoelectronic.
In order to compare the sizes of atoms and ions they have to be isoelectronic: i.e.have the same electron configuration. In this case, each has the [Ne] electronicconfiguration, but a different number of p+ attracting the same number of e−. The− ion will be largest, then the atom, then the + ion.
Answer: Na+ <Ne< F− or F− >Ne>Na+
Exam
ple.
34
PERIODICITY Periodic trends 3
Melting and boiling points
Alkali metals metallic bond strength increases when the charge density increases:smaller atoms with more valence electrons. The number of valence electrons is thesame for the alkali metals, and charge density increases with smaller atoms. Liispredicted to have the highest MP and BP.
Halogens are apolar molecules with only London dispersion forces between them,which increase with increasing molecular mass. I2 is predicted to have the highestMP and BP.
The melting and boilingpoints (MP and BP) ofcompounds criticallydepend on the bondstrength between theparticles, so thissection should beunderstood afterbonding is introduced.
Chemical reactivities
Alkali metals all contain 1 electron in their valence shell (s1); in order to attain thenoble gas configuration these elements react by donating an electron. The easier itis to remove the electron, the more reactive the element. Cs is most reactive and Lileast, since all other things equal the e− in Cs is furthest away from the nucleus.
Halogens all contain 7 electrons in their valence shell (s2p5); in order to attain the noblegas configuration these elements react by accepting an electron. The stronger theattractive force to electrons, the more reactive the element. F2 is most reactive andI2 least, since all other things equal the e− in F2 are closer to the nucleus.
Although the more reactive alkali metals and halogens will react more vigorously, theywill undergo the same chemical reactions. You have to learn some of these reactions byheart, but fortunately all alkali metals and halogens undergo the same reactions and sothey are interchangeable.
Type Equation
Alkali metals+ water
2Li+ 2H2O−−→ 2LiOH+H22Na+ 2H2O−−→ 2NaOH+H2. . .
forms alkaline solutionsand H2 which cancombust (violently)
Alkali metals+ halogens
2Na+Cl2 −−→ 2NaCl2K+F2 −−→ 2KF. . .
forms ionic halides,more reactive alkalimetals +more reactivehalogens produce amore vigorous reaction
Displacementreactions
2KBr+Cl2 −−→ 2KCl+Br2 the more reactivehalogen displaces theions of the less reactivehalogenIonic
reactions
2Br−+Cl2 −−→ 2Cl−+Br22I−+Cl2 −−→ 2Cl−+ I22I−+Br2 −−→ 2Br−+ I2
Table 3.1: Reactions involving Halogens and Alkali Metals
35
PERIODICITY Transition elements
3.3 Transition elements
Transition elements an element whose atom has an incomplete d sub-shell,or which can give rise to cations with an incomplete d sub-shell. Ions oftransition elements have characteristic properties due to thier partiallyfilled d-subshell.
Zn, though part of the d-block, is not a transition metal. The common oxidation state isZn2+ which does not have a partially filled d-subshell. Instead, both Zn and Zn2+ have acompletely filled d-subshell.
atom ion
Zn [Ar]4s23d10 Zn2+ [Ar]3d10
The result is that Zn does not show the characteristic properties of transition metals thatarise from their partially filled d-subshell.
Variable oxidation states
Every transition metal can form ions with charge +2 owing to the fact that 4s electronsare lost before any 3d electrons. But many of the transition metals can also occur inother oxidation states: e.g. Fe2+ and Fe3+, Cu+ and Cu2+.
Complex ions
Complex ion a number of ligands formdative covalent bonds to a positive ion
Ligand a molecule or anion that donates anon-bonding pair of electrons to form adative covalent bond with a metal ion
AlOH2
OH2
OH2
OH2
OH2
OH2
3+
Coordination number the number of ligands around the central ion
Most of the transition metal ions and some non-transition metal ions form hexahydratedcomplexes in water, such as [Al(H2O6]
3+. Six water molecules donate one electron pair
36
PERIODICITY Transition elements 3
each, forming six dative covalent bonds. By applying the VSEPR theory from the nextchapter, the central ion is surrounded by 6 electron domains so these complexes with acoordination number of 6 have an octahedral geometry.
Complexes do not only form between ions and water molecules, but can form withmany other substances so long as the ligand can donate an electron pair (lewis base). E.g.ammonia readily displaces water ligands, because it can form stronger dative covalentbonds than water.
The order of ligand bond strength is called the spectrochemical series:
I− < Br− <Cl− < F− <OH− <H2O<NH3 <CN−
3.3.1 Partially filled d-subshell
In a free ion, the energy levels of each of the 5 d-orbitals are the same, they are said to bedegenerate. But when bonds with a particular geometry form such as described above,the d-electrons closer to the ligands will have a higher energy than those further away,which results in the d-orbitals splitting in energy. This splitting is what gives transitionmetals their characteristic properties.
3d
free Fe2+
3d
3d
[Fe(H2O)6]2+
3d
3d
[Fe(CN)6]4–
E
Note that the CN –
ligands cause a largersplitting of the d-orbitalenergy level becausethey form strongerbonds
37
PERIODICITY Transition elements
Colour of transition metal complexes
In the previous chapter it was shown that electrons can absorb light, thereby movinginto an excited state. When light is absorbed, an electron moves from a d-orbital withlower energy to a d-orbital with higher energy. The energy difference between the splitd-orbitals is typically in the range of visible light.
[Fe(H2O)6]2+ excited state [Fe(CN)6]
4– excited stateE
Note that an electron,represented by � and �,moved from a d-orbitalwith lower energy to ahigher energy d-orbital
So while [Fe(H2O)6]2+ may absorb red photons, [Fe(CN)6]
4– will absorb photons withmore energy such as blue light, because the energy levels are split further apart. Thecolour that is observed is the complementary colour to the light that is absorbed, so asolution containing [Fe(H2O)6]
2+ will appear green.
The colour of light absorption (i.e. the amount of splitting of the d-orbitals) thusdepends on the identity and oxidation state of the central ion, the identity of the ligandand the coordination number / geometry around the central ion.
Magnetic properties of transition metals
When orbitals are partially filled, often (some of) the electrons in the orbitals areunpaired. Each electron has a spin, which when two electrons are paired cancel eachother out. When electrons are unpaired each produces a tiny magnetic field. When thosespins line up in a material they can produce a substantial magnetic field.
38
4BONDING
The most stable electron configuration for each element is when thevalence (= outermost) shell is completely occupied; this is known as the noble gasconfiguration. In order to attain the noble gas configuration atoms will share e− (incovalent and metallic bonds) or exchange e− (forming ions, and ionic bonds).
Depending on which element types combine we can predict the type of bonding:
substance
metal ionic compound molecular compound
only metalelements
metal + nonmetalelements
only nonmetalelements
Electrostatic forces the attractive and repelling forces between charges.
Charge density the ratio between the charge of a particle compared to itssize. The greater the charge density is, the greater the electrostaticattraction. Charge density increases when:
– the charge difference is greater, or when– the distance is smaller between the charges (smaller atoms/ions).
When whole chargesattract, such asbetween ions, theelectrostatic forces willbe strong; and whenpartial charges attract,such as between δ−
and δ+, theelectrostatic forces willbe weak
This chapter explores how physical (macroscopic) properties of a substance are aconsequence of the bonding and arrangement of particles (microscopic properties).
Physical properties
melting point (MP) & boiling point (BP): increases as bond strength increasessolubility: like dissolves likeconduction: requires charged particles that can move to carry current.malleability / ductility: depends on how well the substance can adapt to a change in
the orientation of its particles.
39
BONDING Metallic Bonding
4.1 Metallic Bonding
The valence electrons of metal atoms become detached from individual atomsforming a “sea of delocalised electrons”, which can move freely through thelattice of metal nuclei.
Metallic bond the attraction due to theattraction of + metal nuclei to thedelocalised “sea of electrons” − thatmoves freely Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
∗
∗
Bond strength: MP & BP
The MP/BP of metals are high becausemetallic bonds are typically strong, due tothe electrostatic attraction between +metal nuclei and − free electrons.
Variation among metals depends oncharge density, which increases:
– when a metal has more delocalisedelectrons, and
– with smaller size of the metal atoms.
Metal Delocalised e− Radius MP
Mg 2 140 pm 650 ◦CNa 1 160 pm 98 ◦CK 1 227 pm 64 ◦C
Solubility
Metals only dissolve in other metals, toform alloys.
Many sources incorrectly state that somemetals dissolve in acid. Instead, the metalreacts with the acid in a redox reaction inwhich the metal is converted into themetal cation, which is soluble in water:
M(s)−−→Mx+(aq)+ xe−
Conduction
All metals can conduct electricity: sincemetals contain charged particles: the ‘seaof delocalised electrons’ that can movefreely. The positive metal ‘ions’ cannotmove in the metallic lattice.
Malleability / ductility
When metals are reshaped, the positive ions in the metallic lattice slide over each other.The ‘sea of electrons’ adapts to the different orientation, maintaining the metallic bonds.
In alloys, the presence of foreign metal atoms disturb the host metal lattice structure,causing it to become harder to slide over each other. Alloys are less malleable.
40
BONDING Ionic bonding 4
4.2 Ionic bonding
Ions form when electrons are transferred from one element to another (redox):– reducing the number of valence electrons, to form a metal cation– increasing the number of valence electrons, to form a non-metal anion
Ionic bonds the attractive electrostatic forcesbetween ions of opposing charge, holdingions in a lattice
Mg2+ O2– Mg2+
O2– Mg2+ O2–
Bond strength: MP & BP
The MP/BP of ionic compounds are highbecause ionic bonds are typically strong,due to the electrostatic attraction between+ cations and − anions.
Variation among ionic compoundsdepends on charge density, whichincreases:
– when the ionic charge is larger, and– with smaller size of the ions.
Solubility
In order to dissolve ionic compounds, theionic bonds in the lattice between the ionshave to be broken and replaced bystronger solvation bonds. The easier theionic bonds are broken, and the strongerthe solvation bonds are, the more solublethe ionic compound is.
Solvation solute ions are surrounded by aconcentric shell of solvent molecules,forming solvation complexes
Hydration is solvation by water molecules
AlOH2
OH2
OH2
OH2
OH2
OH2
3+
Conduction
All ionic compounds contain ions, whichare charged particles. But ions in a solid (s)lattice cannot move, and so not conductelectricity. Only when the ioniccompound is liquid (l) or dissolved (aq)can the ions move freely, to conductelectricity.
Malleability
When one row of ions in the ionic latticemoves over 1 place with respect to anotherrow, then all attractive forces betweenions of opposing charge become repellingforces between ions of the same charge.This leads to a crack in the material,meaning that ionic compounds are brittle.
In solution, conductionis proportional to theconcentration of ions.
41
BONDING Ionic bonding
4.2.1 Ionic compounds
Ionic compound positive metal ions and negative non-metal ions form astrong lattice. Ionic compounds are charge neutral, so the empiricalformula of ionic compounds balances the amount of positive andnegative ions.
Writing the empirical formula for ionic compounds involves balancing the total numberof positive and negative charges. Individual ions have a charge, while ionic compounds asa whole are neutral. The formula of ionic compounds is written as the smallest ratiobetween ions (empirical formula) so that the net charge is neutral (e.g., Na2O).
Polyatomic ions ions containing more than one element
Ammonium: NH +4
Acetate: CH3COO–
Cyanide: CN–
Hydrogen carbonate: HCO –3
Hydroxide: OH–
Nitrate: NO –3
Nitrite: NO –2
Carbonate: CO 2–3
Sulfate: SO 2–4
Sulphite: SO 2–3
Phosphate: PO 3–4
Memorize the namesand formulae of thepolyatomic ions byheart! Write empirical formula of ionic compounds
Write the ionic formula of aluminium sulphite
1. Write down the individual ions (from
memory!)
Al3+ and SO 2 –3
2. Determine the smallest possible ratio of
combining the ions to form a neutral
compound. Trick: cross multiply the
charges
charges 3+ and 2−, so 2× 3+ and
3×−2
3. Write the ionic formula, ratio in subscript.
Put brackets (· · · ) around polyatomic ions
if more than 1. Do not put charges!
Al2(SO3)3 using brackets around SO 2 –3
because there is more than one
42
BONDING Covalent bonding 4
4.3 Covalent bonding
Nature of covalent bonding: “shared electron pair”
In molecules, non-metal atoms share electrons to achieve the noble gas configuration.
Hydrogen has one electron in its outer shell, so it needs one more in order to achieve theHe noble gas configuration. Two hydrogen atoms form a single covalent bond: they bothdonate their 1 available valence electron to form a single shared pair. The shared pair ofelectrons lies in the region between the two nuclei and is attracted to them both.
While hydrogen achieves the noble gas configuration with 2 electrons in its outer shell,other non-metal atoms require 8 e− in their valence shell to achieve the noble gasconfiguration. To determine bonding of those elements we use the octet rule.
Bond strength
Covalent bonds form due to the electrostatic attraction of two + atomic nuclei to ashared electron pair, and since the electrostatic attraction between whole charges isstrong, covalent bonds are also strong bonds (on par with metallic and ionic bonds).
In the table below, the bond strengths and lengths of single, double and triple bonds arecompared. Stronger bonds are shorter. Note that a double bond is less then twice asstrong as a single bond.
Bond energy Length
Single C−C 346 kJ mol−1 0.154 nm
Double C−−C 614 kJ mol−1 0.134 nm
Triple C−−−C 839 kJ mol−1 0.120 nm
+268
+225CH3 CH CH2
0.154 nm 0.134 nm
Nomenclature of simple molecular compounds
I The number of atoms is prefixed by a Greek numeral prefix.1. mono 2. di 3. tri 4. tetra 5. penta
II The more electronegative atom is written lastly and is suffixed by -ide.
Molecular formula Systematic name Trivial name
H2O dihydrogen oxide waterCO2 carbon dioxideCO carbon monoxideS2Cl2 disulfur dichloride
Exam
ple.
43
BONDING Covalent bonding
Coordinate covalent bond = dative bond
Dative Bond one atom donates a non-bonding pair forming a covalent bond(instead of both of the atoms donating 1 electron). Such a bond can beindicated by an arrow→.
CO NH4 NH3BF6 Al2Cl6
C ON⊕
H
H
H
HN⊕
H
H
H
B
F
F
F
Cl⊕
Al
Cl
Cl
Cl⊕
Al Cl
Cl
Exam
ple.
4.3.1 Lewis structures
Bonding electron pair pair of valence electrons that are shared betweenatoms (involved in a covalent bond).
Non-bonding electron pair / lone pair pair of valence electrons that are notshared between atoms.
Lewis structures shows all valence electrons on every atom, bonding andnon-bonding, using either lines, crosses, dots or a combination
Octet rule atoms tend to form a stable arrangement with 8 electrons in theirvalence shell= 4 e– -pairs, corresponding to the noble gas configuration.
Only line notation isused in this study guide
Importantly, some elements never deviate from the octet rule while others regularly do.This has to do with the max. number of electrons that are allowed in the valence shell:first period elements (H) in the periodic table have no more than 2 valence electrons, andsecond period elements (B, C, N, O, F) can never have more than 8 valence electons.
But elements from the third period and further can have expanded octets, meaning thatthey have more than eight electrons around one atom. This is possible because the3rd-shell can have a total of 18 electrons. To summarize:
– H will have max. 2 electrons in its valence shell = surrounded by 1 e– -pair– B will oftentimes have only 6 electrons in its outer shell = 3 e– -pairs.– C, N, O & F will always obey the octet rule.– P, S, Cl, Br & I and other non-metal elements beyond the 3rd period prefer to have
an octet, but can expand their octet to have up to 18 electons in their valence shell.
44
BONDING Covalent bonding 4
How to draw Lewis structures
Calculate the total number of valence electrons → electron pairs.
– derive the number of valence electrons from the group number of the element;– add the charge of anions or subtract the charge of cations.
Draw the most likely structure of the molecule.
– the central atom is usually the least electronegative atom;– refrain from drawing cyclic compounds, unless explicitly stated.
Add lone pairs to the outer atoms, until each has an octet (or duets for H).
– place remaining lone pairs on the central atom (expanded octet);– if the central atom has fewer electrons than an octet, exchange lone pairs from the outer
atoms to form double or triple bonds to the central atom to achieve an octet;– resonance occurs when the double bond can be placed in multiple equivalent positions.
Determine the formal charge (FC) on each atom separately.
– formal charge= 0 when atoms have their regular number of bonds, e.g. H: 1 bond,C: 4 bonds, N: 3 bonds, O: 2 bonds, F: 1 bond etc.
– formal charge= number of valence electrons− 1× number of bonds− 2× number of lone pairs
– the formal charges should add up to the overall charge of the molecule or ion.
1
2
3
4
Draw the Lewis structure of XeF6.
1
2
3
4
5
Number of valence electrons of Xe= 8 and F= 7
Draw the most likely structure of the molecule.Xe has lower electronegativity than F
Make octets on the outer atoms and add theremaining lone pair(s) to the central atom.
The formal charge of each atom is 0.
The octet is already expanded.
8 + 6 · 7 = 50ve−
or 25 e– -pairs
Xe F
F
F
F
F
F
Xe F
F
F
F
F
F
Exam
ple.
45
BONDING Covalent bonding
Resonance structures
When multiple Lewis structures can be drawn that satisfy the above rules, electrons aredelocalised which contributes to the stability of a molecule or polyatomic ion.
Draw the Lewis structure of O3 and show the resonance structures and theresonance hybrid.
1
2
3
4
5
The number of valence electrons of O= 6
Draw the most likely structure of the molecule.
Make octets on the outer atoms and add theremaining lone pair to the central atom.
Form an octet on the central atom by exchanging alone pair for a double bond.
Determine the formal charge on each atom.
The octet cannot be expanded.
3 · 6 = 18ve−
or 9 e– -pairs
OO O
OO O
OO O
⊕O
O
O
In step 3 the placement of the double bond on the left or right of the central atom isarbitrary. The real structure of ozone has an electron distribution that is the mean ofthe two possible structures, which can be represented by the resonance hybrid.
⊕O
O
O
⊕O
O Oor
OO O
(0)
resonance hybrid
The resonance hybrid shows the distribution of the double bond over the twoO O bonds. Resonance hybrids must always be drawn with: 1. square bracketsaround them, 2. without any lone pairs and 3. without the formal charges. And 4. ifthe overall charge is not 0 it must be indicated on the top right.
Exam
ple.
46
BONDING Covalent bonding 4
4.3.2 VSEPR: shapes of molecules and ions
Electron domain every bonding or non-bonding pair surrounding an atom.Single, double and triple bonds count as 1 electron domain, because theyare situated between two atoms in the same ‘spatial domain’
VSEPR (Valence Shell Electron Pair Repulsion Theory) electron domainsarrange around the central atom to be as far apart from each other aspossible.
The shape of molecules and ions can be predicted by accounting the number of electrondomains. The geometry of the central atom is a contraction of the number of electrondomains, and the number of bonding vs. non-bonding electron pairs. To determine theshape of a molecule you must always first draw the Lewis structure!
Non-bonding electron domains have a stronger repulsion than bonding-electrondomains. When a molecule has a non-bonding pair, it pushes the bonding pairs slightlycloser together, resulting in decrease of 2.5° per non-bonding pair.
Figure 4.1
C
H
H
H
H
109.5°methane
NH
H
H
107°ammonia
OH
H104.5°water
Methane, ammonia and water each have tetrahedral electron domain geometry. But withincreasing number of lone-pairs, they have different molecular geometries.
47
BONDING Covalent bonding
Electron domain geometry the 3d arrangement of atoms and lone pairs.
Molecular geometry the 3d arrangement of only atoms.
In the table below the central atom is draws as and the surrounding atoms as . The
lobe with two dots represents a lone pair.
2 Electron Domains: linear ED geometry
Bondingpairs
Non-bondingpairs
Name of the molecular shape(atoms only) Bond angle Molecular shape
2/4 0 linear 180°
3 Electron Domains: trigonal planar ED geometry
Bondingpairs
Non-bondingpairs
Name of the molecular shape(atoms only) Bond angle Molecular shape
3/4 0 planar triangular 120°
2/3 1 bent / v-shape 117.5°
48
BONDING Covalent bonding 4
4 Electron Domains: tetrahedral ED geometry
Bondingpairs
Non-bondingpairs
Name of the molecular shape(atoms only) Bond angle Molecular shape
4 0 tetrahedral 109.5°
3 1 trigonal pyramid 107°
2 2 bent / v-shape 104.5°
49
BONDING Covalent bonding
5 Electron Domains: trigonal bipyramidal ED geometry
Bondingpairs
Non-bondingpairs
Name of the molecular shape(atoms only) Bond angle Molecular shape
5 0 trigonal bipyramidal 90°/120°
4 1 see-saw 90°/117.5°
3 2 T-shape 90°
2 3 linear 180°
50
BONDING Covalent bonding 4
6 Electron Domains: octahedral ED geometry
Bondingpairs
Non-bondingpairs
Name of the molecular shape(atoms only) Bond angle Molecular shape
6 0 octahedral 90°
5 1 square pyramid 90°
4 2 square planar 90°
51
BONDING Intermolecular forces
Polar bonds: from covalent to ionic bonding
Ionic bonds form when atoms transfer electrons, covalent bonds form when atoms shareelectrons. If only life was that simple! In reality, electrons of many covalent bonds arenot shared equally: elements with a greater electronegativity (EN) have a greater pull onthe shared electron pair: resulting in a polar bond.
See chapter 3 for thedefinition ofelectronegativity
The element that has higher EN pulls e− closer and so gains a partial negative charge(δ−), leaving a slightly electron deficient atom with a partial positive charge (δ+).
difference between electronegativities (∆ EN)
0 ≤ 0.4 ≤ 1.8
non-polar polar ionic
So the∆EN between two elements determines whether the bond will be:
pure covalent: ∆ EN ≤ 0.4 (approx. equal sharing)polar covalent: 0.4<∆ EN ≤ 1.8 (unequal sharing)ionic: ∆ EN > 1.8 (no longer sharing, e− are transferred to form ions)
Whether a bond is covalent or polar covalent inside a molecule, determines the type ofbonding between molecules→ intermolecular forces dictate the properties of molecularcompounds.
4.4 Intermolecular forces
The physical properties of (simple) molecular compounds are determined by thebonding between molecules. There are three main types of bonding between molecules(= intermolecular bonding):
1. hydrogen bonding (strongest & most polar)2. dipole-dipole bonding (weaker & still polar)3. Londen dispersion forces (weakest & non-polar)
52
BONDING Intermolecular forces 4
4.4.1 Hydrogen bonding
Hydrogen bonding a directional and medium strength bond betweenmolecules that occurs when a molecule contains (−OH,−NHx or HF).
When a H-atom is bound directly to a small very electronegative atom (O, N or F), theshared e− pair is strongly pulled toward the electronegative atom. Since H has no innershell electrons, the positive nucleus is exposed more than it would be for other elements.The larger unshielded charge allows the formation of hydrogen bonds.
The more H-bond forming groups a molecule has (relative to the size of the molecule),the more H-bonds can be formed and so the stronger the IMF, thus the higher MP/BP.This is illustrated in Table 4.1.
Table 4.1
1,2-ethanediol ethanol 1-propanol
Structure HO C
H
H
C
H
H
OH C
H
H
H
C
H
H
OH C
H
H
H
C
H
H
C
H
H
OH
BP 197.3 ◦C 78 ◦C 97 ◦CMP −12.9 ◦C −114 ◦C −126 ◦C
4.4.2 Dipole-dipole forces
Dipole molecule a molecule with asymmetrically distributed partialnegative and positive charge
– Presence of a polar bond: ∆ EN > 0.4– Asymmetric distribution of δ+ and δ+
Dipole-dipole bond electrostatic attraction between the partial charges oftwo dipole molecules
Hδ+
Clδ−
Hδ+
Clδ−
Ion-dipole bond electrostatic attraction between an ion and the partialcharge of a dipole molecule
Hδ+
Clδ−
Clδ−
Hδ+
Na+
53
BONDING Intermolecular forces
Identifying dipole molecules
H2O H2CO CO2
1. Draw the Lewis structure H O
H CH
HO
O C O
2. Polar bonds?
(0.4<∆EN< 1.8)
OH: 3.5− 2.1= 1.4 CH: 2.5− 2.1= 0.4CO: 3.5− 2.5= 1.0
CO: 3.5− 2.5= 1.0
3. Draw the 3D molecule
Oδ−
Hδ+
Hδ+
Cδ+
H
HOδ− δ−O C
δ+
Oδ−
4. Geometric charge distribution? No: dipole molecule No: dipole molecule Yes: not a dipole
molecule
Bond strength
Electrostatic attraction increases as the charge difference increases. So:– ion-dipole > dipole-dipole > dipole-induced dipole– polar bonds with larger∆EN> less polar bonds with smaller∆EN
4.4.3 London (dispersion) forces
London Dispersion (LD) forces electrons oscillate in all molecules, causinga temporary instantaneous dipole. This in turn can induce a temporaryinstantaneous dipole in a neighbouring molecule, resulting in a weakelectrostatic attraction between the molecules
Never use the term‘van der Waals forces’,which is a collectiveterm for several forces,because the IB is verystrict about naming theprecice type of bondingthat occurs.
Bond strength
The LD forces increase as the polarisability of molecules increases:– as the number of electrons increases = the mass of a molecule increases.
Heavier molecules tendto have more electrons,so it is common to saythat the strength of theLondon dispersionforces increases withmolecular mass.
– as the surface area of a molecule increases. Branched molecules have a smallersurface area, are less polarizable, so they experience smaller LD forces.
mass BP
F2 38 u −188 ◦C
Cl2 71 u −34 ◦C
Br2 160 u 58 ◦C
I2 254 u 183 ◦C
mass BP
CH3 CH2 CH2 CH2 CH3 72 u 36 ◦C
C CH3
CH3
CH3
H3C 72 u 10 ◦C
54
BONDING Properties of molecular compounds 4
4.5 Properties of molecular compounds
Note that the properties of simple molecular compounds are very different from giantcovalent compounds. The properties of (simple) molecular compounds depends on thetype of attractive forces between the molecules, in order of decreasing strength: hydrogenbonding > dipole-dipole > London dispersion forces.
Propane Ethanal Ethanol
molar mass 44 g mol−1 44 g mol−1 46 g mol−1
structure CH
H
H
C
H
H
C
H
H
H H C
H
H
δ+
C
δ−
O
H
H C
H
H
δ+
C
H
H
δ−
Oδ+
H
boiling point −42.2 ◦C 20.8 ◦C 78.5 ◦C
bonding type
London dispersiondipole-dipole
London dispersion
hydrogen bondingdipole-dipole
London dispersion
polarity non-polar polar polar
Tip: when asked whattype of bonding ispresent all types mustbe given. So do notforget about LD-forces,as they are alwayspresent.
Melting-/boiling point
Comparing molecules of approximatelythe same molar mass (constant Londondispersion forces); molecules that formhydrogen bonds will have highest MP/BP,then dipole molecules, then moleculesthat only experience LD forces.
Note: the weaker the intermolecularforces, the lower the MP/BP and themore volatile the substance.
Solubility
Like tends to dissolve like: so polarcompounds dissolve well in polar solvents(e.g., water), and nonpolar compounds innonpolar solvents (e.g., oil, hexane).
Conduction
The requirements for conduction are thatthere are charged particles which can move.Molecular compounds are never charged,and so can never conduct electricity. (Butions that are dissolved in tap water canconduct electricity)
Malleability
Whether molecular compounds are malleable depends on the type, directionality andstrength of intermolecular forces. Generally speaking, hydrogen bonds and giantcovalent bonds are highly directional, so solids wherein these play a key-role are notmalleable. Nevertheless, a huge range between brittle (e.g., glass) and malleable(e.g., clay) exists.
55
BONDING Properties of molecular compounds
4.5.1 Giant covalent structures
Giant covalent structures all the atoms are bonded by covalent bonds in acontinuous network extending throughout the material
Allotropes occur when an element can exist in different crystalline forms.
In a network solid there are no individual molecules,and therefore no intermolecular forces. The entiresolid may be considered one big (macro)molecule.The chemical formula is the simplest ratios of thecomponent atoms (empirical formula).
Examples of giant covalent structures are diamondand graphite (discussed below), Si (which is likediamond) and SiO2 known as silica and quartz. Sinceevery O-atom is shared between two Si-atoms, wecan say that every Si-atom is bonded to four halves ofan O-atom; in other words that for every one Si-atomthere will be two O-atoms.
Structure of SiO2
Si O Si
O
O
OSi
O
O
O
Si O SiOSi
O
Si O SiOSi
allotropes of C Diamond Graphite Fullerene
macrostructure network stacked flat layers ofhexagonal C lattices
hexagonal C latticefolded in on itself
shape tetrahedral trigonal planar trigonal planar
bond angles 109.5° 120° < 120°
bonds single bonds 1× double bond 1× double bond
conduction no, e− cannot move yes, delocalised e−
can move betweenlayers
yes, delocalised e−
can move oversurface
56
BONDING Molecular orbitals 4
4.6 Molecular orbitals
Atomic orbitals fuse to form new ‘molecular orbitals’:
σ bond is formed when two atomic orbitals overlap head-on (along a linedrawn through the two atoms). Single or first bond between atoms.
π bond is formed when two atomic orbitals overlap sideways (forming tworegions of electron density on opposite sides of the σ bond).Second/third bond between atoms.
s s
s p
p p
σ σ
π
σ bond
σ and π bond
4.7 Hybridization: mixing atomic orbitals
Hybridization the mixing of atomic orbitals of different energy to obtainhomogeneous orbitals of the same (degenerate) energy
The atomic s and p orbitals mix to form new hybrid orbitals, to form covalent bondswith other atoms. The shape and energy level of the new orbitals are a mix of theconstituent s and p atomic orbitals.
57
BONDING Hybridization: mixing atomic orbitals
In the image on the side, the atomic 2s and 2p subshells of Care not hybridized. From it we cannot deduce why carbon hasbond angles of 109.5°. In order to understand the bondgeometry of C we have to mix atomic orbitals:
– mix one s-orbital with three p-orbitals to get fourdegenerate sp3 orbitals.
– mix one s-orbital with two p-orbitals to get threedegenerate sp2 orbitals and one p-orbital.
– mix one s-orbital with two p-orbitals to get twodegenerate sp orbitals and two degenerate p-orbitals.
non-hybridized C
Unhybridised sp3 hybridised sp2 hybridised sp hybridised
1s
2s
2p
E
1s
sp3
1s
sp2
2p
1s
sp
2p
Bond type only single 1× double 1× tripleor 2× double
ExampleH
H
H
H
H
H
H
H
H H
Basic geometry tetrahedral planar triangular linearBond angles 109.5° 120° 180°
58
BONDING Ozone and oxygen 4
4.8 Ozone and oxygen
Photodissociation chemical bonds absorb enough light energy to break.
Radical a highly reactive molecule that contains an unpaired electron.
Ozone absorbs UV light, protecting us from its ionizing effects (UV light possessessufficient energy to separate an e– from molecules, which can cause e.g. skin cancer).When the atmosphere contains less ozone, less UV light is absorbed.
When the formation of ozone equals the depletion of ozone, there is no net change (=steady state). Human-made pollutants such as CFCs and nitrogen oxides catalyse thedepletion of zone, and have disrupted the steady state.
Formation of ozone
Ozone is continuously produced naturally in the stratosphere by the photodissociationof oxygen. The strong O−−O bond requires high energy UV light to break the bond,forming two O atoms. The O atoms will react with O2 molecules forming ozone O3.
O−−O(g)high energy UV−−−−−−−−→ 2O(g)
O(g)+O2(g)−−→O3(g)
Ozone protection mechanism
As discussed in section 4.3 covalent bonding; double bonds are stronger than singlebonds. And the strength of the resonant bonds in ozone lie between that of a single anddouble bond (bond order 1.5). Remember:
bond strength & lengthare equal due toresonance.
O O
oxygen
⊕O
O
O
⊕O
O O
ozone resonance structures
OO O
(0)
ozone resonancehybrid
To break a O O bond in O3, less energy is required than to break the O O bond inO2. So while oxygen can absorb high energy UV light with wavelengths < 240nm,ozone absorbs wavelengths < 390nm, thus offering protection against medium energyUV light with wavelengths between 240 nm to 390 nm.
59
BONDING Ozone and oxygen
When O3 absorbs light, one ozone bond breaks forming O2 and an O atom:
O3(g)medium energy UV−−−−−−−−−−→O(g)+O2(g)
Ozone is regenerated if the O atom (which is reactive due to it not having a stable octet)collides with an O2 molecule – which probability is much larger than two O atomscolliding to form O2.
O(g)+O2(g)−−→O3(g)
Depletion of ozone
Radicals form due to the homolytic fission of a covalent bond, leaving the two fragmentsboth with an unpaired electron. The radical can wreak all kinds of havoc during variouspropagation steps, whereby a radical reacts with a non-radical forming a non-radical anda radical. Finally, when to radicals collide these can form a pair again, terminating bothradicals.
When CFC molecules such Freon-12 (CCl2F2) enter the stratosphere radical initiationoccurs by absorbing UV light. For many propagation cycles the radicals will react withO3.
It is estimated that a single CFC molecule on average breaks down 105 O3 molecules.
Similarly nitrogen oxides decompose ozone by a (different) radical reaction mechanism.Although the mechanism is too complex, the overall reaction can be written as (withNOx as a catalyst for ozone breakdown):
2O3 −−→ 3O2
60
5ENERGETICS
5.1 Temperature vs heat vs enthalpyTemperature T a measure of the average kinetic energy of particles in a
substance(= independent of the amount of substance)
Heat Q the amount of thermal energy in a given amount of substance(= proportional to the amount of substance)
Heat is a form of energy exchange, its direction depends on the difference intemperature. Heat flows from a warmer to a cooler substance.
Enthalpy H the amount of chemical potential energy contained in one molof a substance, stored in its chemical bonds.
Read “enthalpy” as“chemical energy”
Higher enthalpy: substance hasweaker bonds, can release more heat.
Lower enthalpy: substance hasstronger bonds, can release less heat.
The absolute value of enthalpy cannot be measured, so we measure theenthalpy change (∆H ) between products and reactants in kJ mol−1.
Enthalpy: chemical potential energy
Chemical energy is a form of potential energy, like gravitational potential energy. Fallingfrom a height of 8 km is deadlier than falling from 1 m, since a lot of gravitationalpotential energy is converted into kinetic energy.
Chemistr
y
IB Academ
y
Mount Everest
8 kmChem
istry
IB Academ
y
Chair
0.5 m
Although the stickman on top of the mountain has a huge amount of potential energy,this has no effect on the amount of his kinetic energy at the top. The kinetic energy (ofhis fall) depends on the height difference.
The amount of heat released or absorbed by a chemical reaction depends on the heightdifference between the initial and final enthalpy. Importantly, the average kinetic energy(temperature) of a system does not depend on the amount of enthalpy!
61
ENERGETICS Temperature vs heat vs enthalpy
Chemical system and the surrounding
It is important to make the distinction between a chemical system and its surroundings.The system consists of a chemical reaction: breaking chemical bonds requires energy(thus cooling the surrounding), while forming chemical bonds releases energy (thusheating the surrounding). Energy is exchanged between the system and its surrounding.
In other words, enthalpy (H ) can be converted into heat (Q), and vice versa. The heatincrease is equal to the enthalpy decrease.
chemicalsystem
(reaction)
surrounding
heat
heat
heat
heat
heat (Q) is converted to enthalpy (H )the surrounding cools down
endothermic
chemicalsystem
(reaction)
surrounding
heat
heat
heat
heat
enthalpy (H ) is converted to heat (Q)the surrounding heats up
exothermic
Measuring enthalpy: calorimetry
To accelerate a train requires much more energy than a pedestrian. Similarly, differentsubstances require different amounts of energy to increase their temperature by 1 K. Toincrease the temperature of 1 kg of water by 1 K requires 4.18 kJ, while iron wouldrequire ten times less energy (0.412 kJ).
Specific heat capacity (c) the amount of heat in kJ (or J) to increase thetemperature of 1 kg (or 1 g) of substance by 1 K.
Note that atemperature change of1 K corresponds to achange of 1 ◦C
Note that thetemperature canalways be input aseither K or ◦C in thisexpression, but thatmass in kg yields kJ,while g will yield J.
The change in the amount of heat is given by:
∆Q = c m∆T
∆Q=heat change [kJ] or [J]c = specific heat cap. [kJ kg−1 K−1] or [J g−1 K−1]m =mass [kg] or [g]∆T = temperature change [K] or [◦C]
62
ENERGETICS Temperature vs heat vs enthalpy 5
In order to measure the enthalpy change of a chemical reaction, we can measure the equalbut opposite heat change of the surroundings. We assume that all the heat released orused by a chemical reaction is used to change the temperature of the surroundings (noheat loss). Furthermore, enthalpy is defined per mol reactant resulting in the followingexpression:
Note: convert heat to kJbefore using thisexpression, since thedatabook and the examwill always use kJ mol−1
as units for theenthalpy change.
∆H =−∆Qsurrounding
nsystem
∆H =enthalpy change [kJ mol−1]∆Qsurrounding=heat change [kJ]nsystem =mole amount [mol]
Calorimetry is a technique to measure the increase or decrease in heat by measuring thetemperature change of the surrounding, typically that of water (c = 4.18 Jg−1 K). Underthe assumption that no heat is lost, the enthalpy change of a chemical reaction can becalculated from its effect on the temperature of its surrounding.
Calorimetry: aqueous reaction Calorimetry: combustion
lid
thermometer
polystyrene
reaction mixture
thermometer
beaker
water
spirit lamp
fuel
The temperature change caused by achemical reaction in aqueous solution ismeasured. Assume that all heat istransferred to the water.
Assume that all the heat that is releasedby the combustion reaction is absorbedby the water.
Determine ∆T from a graph
Time
Tem
pera
ture
24
25
26
27
28
29
30
31
32
reaction starts
∆T
Extrapolate the graph up tothe point when the reactionstarts. This point representsthe maximum temperature ifthe reaction had beeninstantaneous (no heat loss).∆T = Tend−Tstart
= 31.5 ◦C− 25.0 ◦C= 6.5 ◦C
Exam
ple.
63
ENERGETICS Energy diagrams
Calculate the enthalpy change ∆H , when the temperature
change ∆T of the surrounding is given
Calculate the enthalpy of combustion for P4 from the given data:
Mass of water g 150.00
Initial temperature ◦C 25.0
Final temperature ◦C 31.5
Mass of phosphorus burned g 0.0500
1. Calculate number of moles of the limiting
reactant n(P4) =m(P4)M (P4)
=0.0500g
4× 30.97gmol−1
= 4.04× 10−4 mol
2. Calculate heat change (∆Q) of the
surrounding
The temperature of the water changes:
∆Q = c m∆T= 4.18 · 150.00 · (31.5− 25.0)= 4.08× 103 J= 4.08kJ
The reaction isexothermic since∆H = negative, whichis logical because allcombustion reactionsare exothermic
3. Use these two values to calculate∆H.
Convert to kJ first, and invert the sign∆H =− ∆Q
n(P4)=− 4.08kJ
4.04× 10−4 mol
=−1.01× 104 kJmol−1
5.2 Energy diagrams
A chemical reaction can use heat from the surrounding�
T ↓�
to break bonds in thesystem (increasing enthalpy). Or the opposite, release heat to the surrounding
�
T ↑�
when bonds are formed (decreasing enthalpy).
In typical chemical reactions both processes occur: bonds are broken and new bonds areformed. And like your bank account, at the end of the month your balance will haveincreased or decreased by some net amount. When the net amount of enthalpy decreases(heat increases) a reaction is said to be exothermic, and when the net amount of enthalpyincreases (heat decreases) a reaction is endothermic.
Exothermic reaction heat is released by reaction to the surroundings�
T ↑�
,the enthalpy H decreases or∆H < 0.
Endothermic reaction heat is used by reaction from the surroundings�
T ↓�
,the enthalpy H increases or∆H > 0.
64
ENERGETICS Hess’s law 5
Exothermic ∆H < 0 Endothermic ∆H > 0
Diagrams
Extent of reaction
Ene
rgy
reactants
products
transition state
Ea
−∆H
Extent of reaction
Ene
rgy
reactants
products
transition state
Ea
+∆H
Sign of ∆H − (negative) + (positive)
Enthalpy ∆H decreases increases
Heat ∆Q increases decreases
Temperature T increases decreases
Bond strength bonds in products are strongerthan the bonds in the reactants
bonds in products are weakerthan the bonds in the reactants
Typical examples bond formation, combustion,acid/base neutralisation, . . .
bond breaking, ionization (!), . . .
5.3 Hess’s law
Hess’s law the enthalpy change (∆H ) of a reaction depends only on thedifference between the enthalpy of the products and reactants, it isindependent of the reaction pathway.
A
B
C
D
AB
BC
C DE
Z
DE
EZ
AZ
Figure 5.1
C(s)+ 12O2 CO(g)
CO2(g)
∆Hx
+ 12O2
−393kJmol−1
+ 12O2
−283kJmol−1
Figure 5.2
The elevation difference AZ equals the sum of allthe individual distances from A to Z . Starting atA, let’s walk over the diagram to Z via the levelsB , C , D , and E .
AZ =−AB +BC +C D −DE + EZ
So whenever an arrow is in the wrong directionsimply change the sign to reverse the arrow in thecorrect direction.
The diagram above in Figure 5.2 shows a typicalexam question. To calculate∆Hx , start on thereactant side (left) and take the long route (viaCO2) in the direction of the product (right). Flipthe direction of the arrows by changing the signof the values.
∆Hx =−394kJmol−1+ 283kJmol−1
=−111kJmol−1
65
ENERGETICS Hess’s law
Another typical application of Hess’s law on the exam is when multiple componentreaction equations and their corresponding∆H values are given, and you have tocalculate the enthalpy change of a reaction that is a composite of the componentreactions.
Enthalpy change from component reactions
Consider the following equations:
2Fe(s)+ 112O2(g)−→ Fe2O3(s) ∆H = x
CO(g)+ 12O2(g)−→ CO2(g) ∆H = y
What is the enthalpy change of the overall reaction below?
Fe2O3(s)+ 3CO(g)−→ 3CO2(g)+ 2Fe(s)
1. Identify the unique compounds in
the given reaction equations
In the first reaction Fe and Fe2O3, and in the
second CO and CO2.
2. Are the unique compounds on the
same side as in the overall
reaction?
Fe and Fe2O3 are on the opposite sides in the
overall reaction. So we invert:
Fe2O3(s)−→ 2Fe(s)+ 112O2(g) ∆H =−x
3. Do the unique compounds have
the correct reaction coefficients?
CO and CO2 have coefficients 3 in the overall
reaction. Multiply by three:
3CO(g)+ 112O2(g)−→ 3CO2(g) ∆H = 3y
4. Add the∆H values to get the
∆H value of the overall reaction ∆H =−x + 3y
Enthalpy change from component reactions
C+O2(g)−→ CO2(g) ∆H =−394kJ
CO(g)+ 12O2(g)−→ CO2(g) ∆H =−283kJ
Calculate the enthalpy change using the given data for: C(s)+O2(g)−→ CO(g)
1. Identify the unique compounds in
the component equations
C in the first, CO in the second reaction.
2. Are the unique compounds on the
same side as in the overall reaction?
C is on the correct side, CO on the wrong side
so invert the equation:
CO2(g)−→ CO(g)+ 12O2(g) ∆H = 283kJ
3. Do the unique compounds have the
correct reaction coefficients?
Yes they already do, so do nothing
4. Add the∆H values to get the∆Hvalue of the overall reaction
∆H =−394kJ+ 283kJ=−111kJ
66
ENERGETICS Hess’s law 5
5.3.1 Standard enthalpy of formation
Standard state (symbol = −◦ ) is the most stable state of an element orcompound under the standard conditions (pressure 100 kPa,temperature 298 K)
(Standard) enthalpy of formation H−◦f
the enthalpy change when one moleof a substance is formed from its elements in their standard states (atstandard conditions).
H−◦f= 0 for: elements in their standard state (e.g. Cgraphite(s), Fe(s), O2(g),
. . . )
Enthalpy of formation isdefined identically, justnot specifically fromtheir standard states
H−◦f of CH4 is: Cgraphite(s)+ 2H2(g)−−→CH4(g)
H−◦f of H2O is: H2(g)+ 12O2(g)−−→H2O(l)
Figure 5.3 shows the standard enthalpy of formation for NO2 on the left, and N2O2 inthe middle. On the right we see the composite energy diagram for the reaction2NO2 −−→N2O4.
12N2+O2 −−→NO2
elements in their standard state0
H−→
NO2
Hf ,NO2
N2+ 2O2 −−→N2O4
elements in their standard state0
H−→
N2O4
H−◦f ,N2O4
2NO2 −−→N2O4
elements in their standard state0
H−→
2NO2
H−◦f ,NO2
N2O4
H−◦f ,N2O4
∆H
Figure 5.3
Using Hess’s law (invert reactant side):∆H =∑
H−◦f,products−∑
H−◦f,reactants
67
ENERGETICS Hess’s law
5.3.2 Bond enthalpy
(Average) bond enthalpy Hb is the amount of energy required to break onemole of the same type of bond, in the gaseous state, (averaged over avariety of similar compounds.)
Hb = 0 for: free gaseous atoms (e.g. H(g), C(g), Fe(g), O(g), . . . )
Bond breaking isalways endothermic
Hb of the C−H bond in CH4 is: 14CH4(g)−−→ 1
4C(g)+H(g)Hb of the O−H bond in H2O is: 1
2H2O(g)−−→H(g)+ 12O(g)
All compounds must bein the gaseous state
CH4 −→ CH3+H
requires 435 kJ mol−1,CH3 −→ CH2+H
requires 444 kJ mol−1.So instead we take1/the number of bondsin the molecule. Figure 5.4 shows the bond enthalpy of the reactants (H2+F2) on the left, and the
product enthalpy (HF) is shown in the middle. On the right we see the composite energydiagram for the reaction H2+F2 −−→HF.
H−H(g) −−→ 2H(g)F−F(g) −−→ 2F(g)
gaseous atoms0
H−→
H2 + F2
H−◦b,F2
H−◦b,H2
H−F−−→H(g)+F(g)
gaseous atoms0
H−→
2HF
H−◦b,HF
H2(g)+F2(g)−−→ 2HF(g)
gaseous atoms0
H−→
H2 + F2
H−◦b,F2
H−◦b,H2
2HF
H−◦b,HF
∆H
Figure 5.4
Using Hess’s law (invert product side):∆H =∑
Hb,reactants−∑
Hb,products
Limitations of average bond enthalpies
– Bond enthalpies require that all compounds are in the gaseous state.– Since average bond enthalpies are obtained by considering a number of similar
compounds containing the type of bond in question, the bond energy in anyparticular compound may deviate.
68
ENERGETICS Hess’s law 5
5.3.3 Enthalpy of combustion
(Standard) enthalpy of combustion H−◦c is the amount of energy releasedfor the complete combustion of one mole of substance in its standard stateand in excess oxygen (at standard conditions).
H−◦c = 0 for: complete combustion products under standard conditions(e.g. CO2(g), H2O(l), . . . )
Combustion reactionsare exothermic
H−◦c of CH4 is: CH4(g)+ 2O2(g)−−→CO2(g)+ 2H2O(l)H−◦c of C6H12O6 is: C6H12O6(s)+ 6O2(g)−−→ 6CO2(g)+ 6H2O(l)
Figure 5.5 shows the combustion enthalpies of the reactants (C2H4+H2) on the left, andthe product (C2H6) is shown in the middle. On the right we see the composite energydiagram for the reaction C2H4+H2 −−→C2H6.
C2H4+H2
H−◦c ,H2
H−◦c ,C2H4
C2H4 +H2 + 3 12O2 −−→
2CO2+ 3H2O
combustion products0
H−→
C2H6
H−◦c ,C2H6
C2H6 + 3 12O2 −−→
2CO2+ 3H2O
combustion products0
H−→
C2H4+H2
H−◦c ,H2
H−◦c ,C2H4
C2H6
H−◦c ,C2H6
∆H
C2H4+H2 −−→C2H6
combustion products0
H−→
Figure 5.5
Using Hess’s law (invert product side):∆H =∑
H−◦c,reactants−∑
H−◦c,products
69
ENERGETICS Energy calculations
5.4 Energy calculations
Enthalpy calculations can be done by using either the enthalpy of formation, bondenthalpy or the enthalpy of combustion. Since enthalpy values are relative to H = 0, andthis is different in all three cases, the three types of values cannot be used in the samecalculation.
Formation Bond Combustion
H = 0 elements at SATP gaseous atoms combustion products at SATP
Definition forming product breaking reactant bonds burning reactant
∆H =∑
H−◦f,prod−∑
H−◦f,react∑
H−◦b,react−∑
H−◦b,prod∑
H−◦c,react−∑
H−◦c,prod
Calculate reaction enthalpy (formation)
Calculate the reaction enthalpy using the standard enthalpies of formation for:
C2H6(g)+ Cl2(g)−→ C2H5Cl(g)+HCl(g)
1. Write the reaction equation C2H6(g) + Cl2(g) −→ C2H5Cl(g) + HCl(g)
2. Find the enthalpy values (DB. 11-13)
remember: when is H = 0?−84 0 −137 −92.3
3. Calculate∑
Hreactant and∑
Hproductseparately
∑
Hreactant = 1×−84+ 1×0=−84kJ
∑
Hproduct = 1×−137+ 1×−92.3
=−229kJ
When calculating∑
Hremember to multiplythe enthalpy valueswith the reactioncoefficients× number of molecules
4. Use the correct formula for∆H =formation:
∑
Hf,prod−∑
Hf,react
bond/combustion:∑
Hreact−∑
Hprod
∆H =∑
Hf,product−∑
Hf,reactant
=−229−−84kJmol−1
=−145kJmol−1
70
ENERGETICS Energy calculations 5
Calculate reaction enthalpy (bond)
Calculate the reaction enthalpy using the bond enthalpies for:
CH4(g)+ 2O2(g)−→ CO2(g)+ 2H2O(g)
1. Write the reaction equation
in structural formulas
H C
H
H
H + 2O O O C O + 2H O
H
2. Find the enthalpy values
(DB. 11-13)
C H: 414 C O: 804O O: 498 O H: 463
3. Calculate∑
Hreactant and∑
Hproduct separately
∑
Hreactant = 4×414+ 2×498= 2652kJ
∑
Hproduct = 2×804+ 2×2×463
= 3460kJ
× bonds per molecule× number of molecules
4. Use the correct formula for
∆H =∆H =
∑
Hb,reactant−∑
Hb,product
= 2652− 3460kJmol−1
=−808kJmol−1
Calculate reaction enthalpy (combustion)
Calculate the reaction enthalpy using the standard enthalpies of combustion for:
CH3COOH(l)−→ CH4(g)+ CO2(g)
1. Write the reaction equation CH3COOH(l) −→ CH4(g) + CO2(g)
2. Find the enthalpy values (DB. 11-13)
remember: when is H = 0?−874 −891 0
3. Calculate∑
Hreactant and∑
Hproductseparately
∑
Hreactant = 1×−874=−874kJ
∑
Hproduct = 1×−891+ 1×0
=−891kJ
CO2 is a product ofcomplete combustion,so H−◦c = 0
× number of molecules
4. Use the correct formula for∆H =formation:
∑
Hf,prod−∑
Hf,react
bond/combustion:∑
Hreact−∑
Hprod
∆H =∑
Hc,reactant−∑
Hc,product
=−874−−891kJmol−1
= 17kJmol−1
71
ENERGETICS Energy cycles
5.5 Energy cycles
Energy cycles (such as the Born-Haber cycle) are another application of Hess’s law,meaning as much as: “start at the reactant enthalpy level and add all the values whilemoving toward the product energy level”. See Figure 5.1 on 65.
Enthalpy of atomization HAT the enthalpy change when one mole ofgaseous atoms are formed from the elements in their standard states.
Cgraphite(s)−−→C(g)12Cl2(g)−−→Cl(g)
Breaking bonds in theelement, and vaporisa-tion/sublimation bothrequire energy, soendothermic.
NB: for gaseous diatomic elements such as O2, Cl2, . . . : H−◦AT =12
H−◦b
Ionization enthalpy HIE the enthalpy change when one mole of gaseousatoms each lose one electron to form one mole of gaseous 1+ ions.
H1stIE: Mg(g)−−→Mg+(g)+ e−
H2ndIE: Mg+(g)−−→Mg2+(g)+ e−
The second ionisationethalpy is actuallydefined as the enthalpychange when one moleof gaseous 1+ ionseach lose one electronto form one mole ofgaseous 2+ ions.
Electron affinity enthalpy HEA the enthalpy change when one mole ofgaseous atoms acquire one electron forming one mole of gaseous 1− ions.
H1stEA: O(g)+ e− −−→O– (g)H2ndEA: O– (g)+ e− −−→O2– (g)
Standard lattice enthalpy H−◦lattice
the enthalpy change when one mole of asolid ionic compound is separated into gaseous ions at standard conditions
NaCl(s)−−→Na+(g)+Cl– (g)Ba2(PO4)3(s)−−→ 2Ba2+(g)+ 3PO 3–
4 (g)
Ionic bonds are fairlystrong, so separatingan ionic solid into itsconstituent (gaseous)ions requires a lot ofenergy. Soendothermic.
NB: lattice enthalpy issometimes defined asthe ionic solid formedfrom its gaseous ions.This inverts the sign ofthe enthalpy, but not itsmagnitude.
Magnitude of lattice enthalpy: depends on the ionic bond strength. Theionic bond strength increases when:
– the ions are smaller (up a group)– the ions have larger charge
72
ENERGETICS Energy cycles 5
Enthalpy of hydration H−◦hydration
the enthalpy change when one mole of
gaseous ions dissolve in water, forming an infinitely dilute solution.
Na+(g)−−→Na+(aq)NO –
3 (g)−−→NO –3 (aq)
Hydration formsion-dipole bondsbetween watermolecules and the ion,so exothermic.
Magnitude of enthalpy of hydration: depends on the ion-dipole bondstrength between water and the ion in question. Since the dipole of water is aconstant here, the ion-dipole bond strength increases when:
– the ions are smaller (up a group)– the ions have larger charge
Enthalpy of solution H−◦solution
the enthalpy change when one mole of solidionic substance dissolves in water, forming an infinitely dilute solution.
NaCl(s)−−→NaCl(aq)
Dissolving can be seenas two seperateprocesses: breaking allthe ionic bonds(H−◦lattice) and formingion-dipole bonds withwater (H−◦hydration).
NaCl(s) NaCl(aq)
Na+(g)+Cl–(g)
∆H−◦solution
∆H−◦lattice ∆H−◦hydration
Using Hess’s law: H−◦solution =H−◦lattice+H−◦hydration
Born-Haber cycles
The amount of energy required to separate one mole of a solid ionic compound intogaseous ions (under standard conditions) is known as the lattice enthalpy. Understandard conditions ionic bonds are too strong to form free gaseous ions, so latticeenthalpies cannot be determined experimentally. Specialized energy cycles based onHess’s law (Born-Haber cycles) are used to calculate the lattice enthalpy.
Let’s study the lattice enthalpy of MgO(s) with the help of figure 5.6. The definition oflattice enthalpy is the enthalpy change when one mole of solid ionic compound isseparated into gaseous ions: MgO(s)−−→Mg2+(g)+O2– (g).
So let’s start at MgO(s) and move toward the free gaseous ions Mg2+(g)+O2– (g), exceptinstead of moving to the gaseous ions directly we use all the other types of enthalpies toarrive at the free gaseous ions.
73
ENERGETICS Energy cycles
H−→
formation of coumpoundfrom elements
separating the crystallattice into gaseous ions
MgO(s)
Mg(s)+ 12O2(g)
Mg(g)+ 12O2(g)
Mg(g)+O(g)
Mg2+(g)+O2– (g)
Mg2+(g)+ 2e– +O(g)
Hf,MgO
HAT,Mg
HAT,O
HIE,Mg
Hlattice,MgO
HEA,O
1
2
3
4
5
elements
atomisation of Mg
atomisation of O
1st+2nd ionisationenthalpy of Mg
1st+2nd electronaffinity of O
1
2
3
4
5
From compound −−→ elements is the reverse of the enthalpy of formation:
From elements −−→ gaseous atoms:
From gaseous atoms −−→ gaseous ions:
−Hf,MgO
+HAT,Mg
+HAT,O
+H1st+2ndIE,Mg
+H1st+2ndEA,O
formation of elements from compound
atomisation of Mg
atomisation of O
ionisation of Mg
electron affinity of O
Figure 5.6
74
ENERGETICS Entropy 5
Using Hess’s law, the enthalpy difference between MgO(s) and Mg2+(g)+O2– (g) equals:
Hlattice,MgO =−Hf,MgO+HAT,Mg+HAT,O+H1st+2ndIE,Mg+H1st+2ndEA,O
Importantly, you’re not expected to learn this formula by heart. You do have to be ableto identify the three steps (reverse of formation, atomisation & ionisation) and predictwhether the steps are expected to be exo- or endothermic.
5.6 Entropy
Entropy S refers to the distribution of the available energy among particles.Nature tends toward an increase in entropy, more ways in which theenergy can be distributed.
S = 0 for: a perfectly ordered crystal at absolute 0 K
Higher entropy: larger disorder,more ways to distribute E,∆S =+
Lower entropy: more ordered, lessways to distribute E,∆S =−
Entropy is also said toquantify the degree ofdisorder or randomnessin a system. Naturetends to an increase inentropy, more disorder.
Increase entropy of a system
1. Creation of a gas disorder increases more during reactions that produce gascompared to the other three factors.
2. Increased number of particles disorder increases when a reaction yields an increasein the number of particles.
3. Change of state solids have least disorder, liquids/solutions have more disorder, andgases have the most disorder.
4. Mixing disorder increases when mixing.
When predictingentropy changes, achange in the numberof gaseous particles (1.)is the deciding factor.
Which reaction has the greatest increase in entropy?
A. 2CH3OH(l)+ 3O2(g)−−→ 2CO2(g)+ 4H2O(l)
B. N2(g)+ 3H2(g)−−→ 2NH3(g)
C. 2HCl(aq)+MgCO3(s)−−→MgCl2(aq)+H2O(l)+CO2(g)
D. NH3(g)+HCl(g)−−→NH4Cl(s)
An increase in entropy corresponds to an increase in disorder. First check reactionswherein the number of gaseous molecules increase:
��ZZA. the number of gaseous molecules decreases��ZZB . the number of gaseous molecules decreasesC . the number of gaseous molecules increases��ZZD. the number of gaseous molecules decreases
Exam
ple.
75
ENERGETICS Entropy
Calculate the entropy change ∆S
Calculate the entropy change using DB. 12 and S−◦H2= 130.7 JK−1 mol−1 for:
CH4(g)+H2O(g)−*)− CO(g)+ 3H2(g)
1. Write the reaction equation CH4(g) + H2O(g) −*)− CO(g) + 3H2(g)
2. Find the entropy values (DB. 12) 186 188.8 197.7 130.7
Notice how thiscalculation is exactlythe same as you’vedone before for theenthalpy of formation.
3. Calculate∑
Sreactant and∑
Sproductseperately
∑
Sreactant = 1×186+ 1×188.8
= 375 JK−1 mol−1
∑
Sproduct = 1×197.7+ 3×130.7
= 590 JK−1 mol−1
× number of molecules
NB: the units of S areJ K−1 mol−1 and NOTkJ K−1 mol−1
4. ∆S =∑
Sproduct−∑
Sreactant ∆S =∑
Sproduct−∑
Sreactant
= 590− 375 JK−1 mol−1
= 215 JK−1 mol−1
5.6.1 Spontaneity
A reaction will occur spontaneously if a system moves from a less stable to a more stablestate. The stability of a system depends on both the enthalpy change and the entropychange. These two factors combine to define the standard Gibbs free energy∆G−◦ :
∆G−◦ =∆H−◦ −T∆S−◦
Standard Gibbs ‘free’ energy change ∆G−◦ quantifies the criterion forpredicting the spontaneity of a reaction: it is related to both theenthalpy and entropy changes.
∆G < 0: spontaneous ∆G = 0: equilibrium ∆G > 0: non-spontaneous
∆G−◦ =∑
G−◦f,products−∑
G−◦f,reactants
76
ENERGETICS Entropy 5
Standard Gibbs ‘free’ energy change of formation ∆G−◦f
quantifies thecriterion for predicting the spontaneity of the formation of acompound from its elements: it is related to both the enthalpy andentropy changes.
∆G−◦f= 0 for elements under standard conditions
Although S values are always positive, the entropy change∆S of a reaction can bepositive or negative. The condition for a spontaneous reaction is as follows:
Enthalpy
exothermic endothermic
∆H < 0 ∆H > 0
Ent
ropy m
ore
diso
rder ∆S > 0 ∆G=∆H −T ∆S
∆G= (−) −T (+)
∆G=∆H −T ∆S
∆G= (+) −T (+)
always (−),so spontaneous
spontaneous if the T∆Sterm is larger than the∆H
term (high T )
less
diso
rder ∆S < 0 ∆G=∆H −T ∆S
∆G= (−) −T (−)
∆G=∆H −T ∆S
∆G= (+) −T (−)
spontaneous if the∆H termis larger than the T∆S term
(low T )
always (+),so never spontaneous
77
ENERGETICS Entropy
Calculate the temperature when a reaction becomes spontaneous
At what temperature does the following reaction become spontaneous?
CaCO3(s)−→ CaO(s)+ CO2(g) ∆H = 179kJmol−1
Substance S−◦ / J K−1 mol−1
CaCO3(s) 92.9
CaO(s) 39.8
CO2(g) 213.7
1. Calculate the change in entropy:
∆S =∑
Sproduct−∑
Sreactant
∆S= (1×213.7+ 1×39.8)− 1×92.9
= 160.6 JK−1 mol−1
× number of molecules
2. Convert to kJ K−1 mol−1
(to match∆H)
= 0.1606kJK−1 mol−1
3. Write the expression for∆Gand equate it to zero.
∆G =∆H−T∆S= 0
4. Rearrange to calculate
minimum value of T .
Substitute∆H and∆S .
T∆S =∆H
T =∆H∆S
=+1790.1606
= 1114K
5. At T greater than the result, the
reaction will be spontaneous.
The reaction will be spontaneous at T > 1114 K
78
6KINETICS
Reaction rate the change in concentration of a particular reactant or productper unit time, measured in mol dm−3 s−1
rate=−∆[reactants]
∆t=∆[products]
∆t
In a concentration vs. time graph, which shows the progress of a chemical reaction, thereaction rate is equal to the tangent or slope. When the slope= 0 (the graph ishorizontal) the reaction has reached either completion or equilibrium.
Completion
A−−→ B
time
[. . . ]
[B]
[A]
time whencompletionis reached
t
reactionrate
Equilibrium
A−−*)−− B
time
[. . . ]
[B]
[A]
time whenequilibriumis reached
reactionrate
Experimental determination
The rate of reaction can be experimentally determined by measuring how theconcentration changes with time. Multiple concentration measurements have to beperformed over time to infer the rate from the change. To measure concentration we can:
– use an absorption spectrometer; absorption is stronger with a higherconcentration (of for example coloured transition metals complexes)
– measure the pH if the [H+] or [OH−] concentrations change– measure the electrical conductivity if the ionic concentration changes– measure the volume of a gas (which is proportional to the amount in mole) that is
evolved from reaction– measure the mass change of the sample as a result of gas formation.
79
KINETICS Collision Theory
6.1 Collision Theory
Collision theory states the three conditions that must be met for a successful reaction totake place:
Collision theory increase rate
1 particles must collide, higher collision frequency2 with proper orientation, —3 and sufficient energy increase collision energy (by increasing T ), lower
energy barrier (Ea)
The collision theory can be used to understand how the rate of reaction can be affected.Every collision with proper orientation and with sufficient energy leads to a chemicalreaction (known as effective collisions). When more collisions occur per second (highercollision frequency), there is a proportional increase in the number of effective collisions.Also, by either increasing the collision energy or lowering the energy barrier, a largerfraction of the collisions are effective. The orientation of collisions is random and cannotbe influenced.
The four factors that increase the rate of reaction:or decrease by doingthe opposite
1. ↑ concentration/pressure: particles are closer together so the collision frequencywill increase.
2. ↑ surface area / ↓ particle size: collisions with solids occur on the surface, soincreasing the surface area will increase the collision frequency.
3. ↑ temperature: particles have a higher amount of average kinetic energy, resultingin an increase in the collision frequency and a larger fraction of the collisions will beeffective by increasing the collision energy.
4. add catalyst: provides an alternative reaction pathway/mechanism which has alower activation energy, so a larger fraction of the collisions will be effective.
Exothermic
Extent of reaction
Ene
rgy
reactants
products
Ea without a catalyst
Ea with a catalyst
transition state
Endothermic
Extent of reaction
Ene
rgy
reactants
products
Ea without a catalyst
Ea with a catalyst
transition state
The activation energy (Ea) is the minimum amount of energy of a collision between twoparticles to lead to a reaction. We can say: the energy required for an effective collision.
80
KINETICS Rate equation and reaction order 6
Ea
(with catalyst)
at lower T
at higher T
(additional) fraction ofparticles that havesufficient energy:
298 K
273 K
at higher T
Kinetic energy
Num
ber
ofpa
rtic
les
Ea Ea
(with catalyst) (without catalyst)
without catalyst
with catalyst
(additional) fraction ofparticles that havesufficient energy:
Kinetic energy
Num
ber
ofpa
rtic
les
6.2 Rate equation and reaction order
The rate equation shows the mathematical relation between the concentrationsof reactants and the rate of the reaction. When A and B react:
Rate equation: rate= k[A]x[B]y
x is the order of the reaction with respect to A, and y with respect to B.
The overall order of reaction: = x + y.
Note: the order of the reaction (and therefore also the rate equation) can onlybe derived from:
1. experimental data or2. from the reaction mechanism if the rate determining step (r.d.s.) is
known.
It cannot be derived from the balanced reaction equation!
81
KINETICS Rate equation and reaction order
Order of a reaction
The order of reaction conveys the mathematical dependence of the rate on theconcentration of the reactant. To simplify things, let’s assume there is only reactant Aand x is the overall order. The rate equation can be written as: rate= k[A]x .
zero order first order second order
rate equation rate= k*1 rate= k[A] rate= k[A]2
[. . . ] vs time
time
[. . . ]
time
[. . . ]
time
[. . . ]
rate vs [. . . ]The rate is equal to theslope of [. . . ] vs time.And note that while thereaction rate decreasesover time (for the firstand second ordergraphs), the greaterconcentration causesthe rate to increase.
[. . . ]
rate
[. . . ]
rate
[. . . ]
rate
units of k*2k = rate
=moldm−3 s−1
k =rate[A]
=moldm−3 s−1
moldm−3
= s−1
k =rate
[A]2
=moldm−3 s−1
(moldm−3)2
=mol−1 dm3 s−1
*1 Since k[A]0 = k*2 The units of the rate constant depend on the overall order of the reaction.
Solving for k yields k =rate[A]x
.
82
KINETICS Rate equation and reaction order 6
Derive the rate equation from experimental data
To derive the order of the reaction, the rate of reaction is measured at varying reactantconcentrations. Keeping all reactant concentrations the same except that of [A], therate∝ [A]x :
zero orderx = 0
first orderx = 1
second orderx = 2
third orderx = 3
double [A] rate∝ 2x rate doesn’t change rate ×21 =×2 rate ×22 =×4 rate ×23 =×8triple [A] rate∝ 3x rate doesn’t change rate ×31 =×3 rate ×32 =×9 rate ×33 =×27
Derive the rate equation from a reaction mechanism
Since the collision of more than two particles with the correct orientation and sufficientenergy is extremely unlikely, many reactions proceed in multiple steps. The step by stepsequence of elementary reactions by which the overall chemical change occurs is calledthe reaction mechanism.
A successful collision between two molecules can lead to an activated intermediate,which can continue to form the product in a subsequent collision, or revert back to theoriginal reactants.
Let’s study the reaction 2A+B−−→A−A−B. A possible mechanism is:
step 1. A+A −−*)−− A−A
step 2. A−A+B −−→ A−A−B
Ene
rgy
A+A
A−A+B
A−A−B
Ea
step 1
Ea
step 2
From the energy diagram we see that the activation energy of step 1. is higher than theactivation energy of step 2. This means that the rate of step 1. will be slower than that ofstep 2. The rate of the overall reaction is determined by the slowest step in a reactionmechanism, known as the rate determining step (r.d.s.). The overall reaction rate willtherefore only depend on the formation of A−A, so the rate= k[A][A] = k[A]2.
83
KINETICS Rate equation and reaction order
A rate equation can be written for every elementary step. The overall rate of reactiondepends on the r.d.s., so when the rate equation of the r.d.s. contains an intermediate itshould be replaced by the (previous) rate equation in which the intermediate is formed.
Mechanism Rate equation Molecularity
step 1. A+Bfast−−→ X rate= k[A][B] bimolecular
step 2. A+Xslow−−→ Y rate= k[A][X] bimolecular
step 3. Yfast−−→ C+D rate= k[Y] unimolecular
overall 2A+Bfast−−→ C+D rate= k[A][X] = k[A][A][B]
Find the rate equation when the reaction mechanism and r.d.s.
are known.
Determine the rate equation given the following mechanism for
2NO+ 2H2 −→ N2+ 2H2O:
step 1. NO+NOfast−*)− N2O2
step 2. N2O2+H2
slow−→ N2O+H2O
step 3. N2O+H2
fast−→ N2+H2O
1. Determine the rate equation of
the r.d.s.
rate= k[N2O2][H2]
2. If the rate equation contains and
intermediate product, replace it
with the [reactants] that form the
intermediate.
N2O2 is an intermediate (it is not present in the
overall reaction), so replace [N2O2] with
[NO][NO].rate= k[NO][NO][H2] = k[NO]2[H2]
3. Repeat step 2 if there is still an
intermediate product
NO and H2 are both reactants in the overall
reaction, no repeat necessary
4. Write down the rate equation rate= k[NO]2[H2]
Arrhenius equation
The rate constant k depends only on the temperature for a particular reaction. TheArrhenius equation relates the rate constant k with the absolute temperature:
k =Ae−EaRT
k = rate constant dependsA = frequency factor dependsEa=activation energy [J]R =gas constant [8.31 J K−1]T = temperature [K]
84
KINETICS Rate equation and reaction order 6
This equation is often written in its logarithmic form, which can be viewed as a linearequation by adapting the x and y-axis:
ln k=−Ea
RT+ lnA
=−Ea
R· 1T+ lnA
= m ·x + c
If ln k is plotted on the y-axis, and1T
on the x-axis:
– then the slope (m) equals−Ea
R– and the intersection with the y-axis (c ) equals lnA, and
So when k is measured at differenttemperatures, and the results are
plotted in a ln k vs1T
graph, then:
– the activation energy can becalculated from the slope
(−Ea
R), and
– the frequency factor A can becalculated fromextrapolation to the ln k-axis.
1T
(K−1)
ln k
gradient=−Ea
R
lnA
You can solve simultaneous equations when you know k at two different temperatures,using the following equation from the databook:
lnk1
k2=−Ea
R
�
1T2− 1
T1
�
85
7EQUILIBRIUM
7.1 Dynamic equilibrium
In a static equilibrium nothing changes, like for example in a mass balance. Chemicalequilibriums are dynamic, there is a constant conversion in both directions such thatthere is no net change.
Dynamic equilibrium the forward rate of reaction equals the reverse rate ofreaction.
Note that the amounts of reactants and products are not (necessarily) equal, theamounts are constant / do not change.
Completion reaction
A−−→ B
time
[. . . ]
[B]
[A]
time whencompletionis reached
Reaction stops when the limitingreagent runs out.
Equilibrium reaction
A−−*)−− B
time
[. . . ]
[B]
[A]
time whenequilibriumis reached
Reaction continues indefinitely butreaches equilibrium.
A system at equilibrium is a mixture with all of the reactants and products present∗, ∗remember: usually notin equal amounts!
andthe concentrations of the reactants and the products do not change. A consequence isthat the macroscopic properties (the colour, viscosity etc.) of the system does notchange. Systems can only remain in equilibrium in a closed system, since the exchange ofmatter with the surroundings would disturb the equilibrium.
87
EQUILIBRIUM Equilibrium law expression
Understanding equilibriums
Using Collision Theory, we know that the reaction rate depends on the concentration.The forward reaction rate is proportional to the [reactant], and the reverse reaction rateis proportional to [product].
reactantrate∝ [reactant]−−−−−−−−*)−−−−−−−−rate∝ [product]
product
At the time that reagents are mixed (t = 0) the forward reaction rate is greatest, but itdecreases over time as the [reactant] decreases. At t = 0 the reverse reaction rate is 0, as[product] = 0. But as [product] increases over time, so will the reverse reaction rate.
At equilibrium the forward reaction rate has decreased and the reverse reaction rate hasincreased to the point that they are equal. It does not matter from which side theequilibrium is approached, at some point the two rates will be equal.
7.2 Equilibrium law expression
Consider the generic chemical reaction given below, in which A+B react to form C+Dand the reaction coefficients are indicated by the small letters pq r s .
pA+ qB−−*)−− rC+ s D
The reaction quotient Q is defined at any point in time during the reaction as:
Q =[C]r × [D]s
[A]p × [B]q=Kc or abstractly as
[products][reactants]
Since the concentrations of the reactants and products do not change at equilibrium, thereaction quotient Q has a very specific value, which is called the equilibrium constantKc (which only depends on temperature).
88
EQUILIBRIUM Equilibrium law expression 7
Determine if a system is at equilibrium / predict direction of shift
to restore equilibrium
Sulphur dioxide reacts with oxygen forming sulphur trioxide in an equilibrium reaction.
At the prevailing temperature all substances are gaseous and Kc = 2. At some point in
time, the concentrations are [SO2] = 2M, [O2] = 1M and [SO3] = 2M.
Determine if the system is at equilibrium and predict direction of the shift to restore
equilibrium.
1. Write down the balanced equilibrium
reaction.
2SO2(g)+O2(g)−*)− 2SO3(g)
2. Derive the reaction quotient Q and
calculate the result. Q =[SO3]
2
[SO2]2× [O2]
=22
22× 1= 1
3. When the equilibrium point is reached
Q = Kc.
Since Q = 1 at the given point in time,
Q 6= Kc, so the system is not in
equilibrium.
4.Use Q =
[product][reactant]
.
If Q > Kc shift to reactant side and if
Q > Kc shift to product side.
Since Q < Kc at the given point in time,
to restore equilibrium the value of Qshould increase by increasing the
[product] / [reactant] , so the system is
not in equilibrium.
Magnitude of Kc =[products]
[reactants]
If Kc has a very large value then the [product]must be much higher then the [reactant].In summary for all values of Kc:
reaction
Kc� 1 (almost) no reactionKc < 1 equilibrium favours reactant sideKc ≈ 1 approx. equal [reactant] and [product]Kc > 1 equilibrium favours product sideKc� 1 tends to completion
89
EQUILIBRIUM Equilibrium law expression
Manipulation of Kc
When an equilibrium reaction is reversed, the equilibrium constant is inversed
(1
Kc). And when chemical reactions are added up together their respective
equilibrium constants are multiplied (Kc1×Kc2).
Reversal: when discussing equilibrium reactions the terms reactant side and productside are often avoided, since it is a matter of perspective. For example, in theequilibrium of 2NO2
−−*)−−N2O4 the following reactions occur at the same time:– 2NO2 −−→N2O4– N2O4 −−→ 2NO2
Both equilibrium reactions 2NO2−−*)−−N2O4 and N2O4
−−*)−− 2NO2 represent thesame equilibrium. What is the forward reaction in the one representation of theequilibrium, is the reverse reaction in the other. The equilibrium law expression ofthe two reactions are each others inverse:
2NO2−−*)−−N2O4 N2O4
−−*)−− 2NO2
equilibrium law expression Kc =[N2O4]
[NO2]2 Kc
′ =[NO2]
2
[N2O4]=
1Kc
Addition: when multi-step reactions are added to give an overall reaction, the increaseof a reaction coefficient cause the power in the quotient Q to increase. This meansthat when reactions are added, the quotients are multiplied:
step 1. A+B −−*)−− X Kc1=[X][A][B]
step 2. A+X −−*)−− C Kc2=[C][A][X]
overall 2A+B −−*)−− C Kcoverall=Kc1×Kc2
= ��[X][C]
[A][B][A]��[X]
=[C][A]2[B]
So the multiplication of the equilibrium expressions indeed leads to the correctequilibrium expression of the overall reaction.
90
EQUILIBRIUM States of matter 7
7.3 States of matter
So far we’ve only considered chemical equilibriums in closed systems, but similarly wecan consider physical states of matter in closed systems. In particular: liquid-vapourequilibrium and precipitation-solution equilibrium of ionic compounds.
Liquid vapour equilibrium Solution equilibrium
Equilibrium H2O(l)−−*)−−H2O(g). NaCl(s)−−*)−−NaCl(aq)
Drawing
liquid
vapourvaporization condensation
Explanation Fast moving particles will escape theliquid phase, evaporating into vapourphase. While slow moving particlescondense into the liquid phase. Whenthe rates of these processes are equal, adynamic equilibrium will establish.
The ability of an ionic compound todissolve depends on the rates ofsolution and precipitation, or in otherwords by the hydration vs. the latticeenthalpies. While table salt dissolvesfully up to the point of saturation, therates of solution and precipitation areat equilibrium after a solution issaturated.
Independent of: – Surface area: affects both evaporationand condensation
– Volume of liquid in container
– Surface area: affects both solutionand precipitation
– Volume of solid in container
Dependent on: – Volume of gas in container(concentration or partial pressure)
– Volume of liquid in container(concentration of dissolved particles)
91
EQUILIBRIUM Le Chatelier’s principle
7.4 Le Chatelier’s principle
Le Chatelier’s principle states how a system that is in equilibrium responds toan external change.
Le Chatelier’s principle when a system at equilibrium is disturbed bychanging the conditions, the system will shift the position ofequilibrium to (partially) counteract the change.
Factors that affect the equilibrium position
The factors that affect the equilibrium position will be exemplified based on thefollowing generic chemical reaction:
3A(g)+B(s)−−*)−− 3C(g)+D(g) ∆H=−195kJmol−1
Stress The position of the equilibrium will: Example
increase [A] shift away from A in order to lower [A] to the product side
decrease [D] shift towards D to partially restore itsconcentration
to the product side
increase P by adecrease in V
shift towards the side with fewer gaseous moleculesto reduce the pressure
so to the reactant side
increase P byaddition of aninert gas
no effect, because the partial pressures of thereactants and products do not change
—
add a catalyst no effect, because the forward and reverse reactionrates are increased equally
—
increase T shift toward the endothermic side (to lower thetemperature) by changing the value of Kc
to the reactant side,since the forwardreaction is exothermic(∆H=−)
92
EQUILIBRIUM Equilibrium calculations 7
7.5 Equilibrium calculations
Completion reaction
– Reaction continues until the limitingreactant is fully consumed.
Equilibrium reaction
– Reaction occurs in both directionssimultaneously.
– At equilibrium the amounts are constant
In equilibrium calculations we always use the RICE table and the equilibrium lawexpression (Kc = . . . ). The concept of limiting reactant is invalid in equilibrium reactions.
Equilibrium calculations
2.0 mol H2(g) and 1.0 mol N2(g) react in equilibrium, forming NH3(g). The volume of the closed
container is 2.0 dm3 and the temperature is kept constant. At equilibrium, 0.50 mol of NH3(g) is present.
Calculate the value of the equilibrium constant Kc.
1. Convert the given values to
concentrations.
Indicate [ . . . ]in or [ . . . ]eq
[H2]in =2.0mol
2.0dm3= 1.0moldm−3
[N2]in =1.0mol
2.0dm3= 0.50moldm−3
[NH3]eq =0.50mol
2.0dm3= 0.25moldm−3
2. Write the balanced equilibrium reaction
and use it as a header for a RICE table.
Change: use x & reaction coefficients
Reaction 3H2(g) + N2(g) −*)− 2NH3(g)
Initial
Change −3x −x +2xEquilibrium
3. Complete the table, use concentrations
Initial: check the text, 0 when unknown
Equilibrium= Initial+ Change
Reaction 3H2(g) + N2(g) −*)− 2NH3(g)
Initial 1.0 0.50 0
Change −3x −x +2xEquilibrium 1.0− 3x 0.5− x 2x(= 0.25)
4. Kc is unknown: use [ . . . ]eq to calculate
x and calculate all values in Equilibrium
Kc is known: insert the expressions
from Equilibrium into Kc = . . .
[NH3]eq is known, so 2x = 0.25 and x = 0.125
[H2]eq = 1.0− 3 · 0.125= 0.625moldm−3
[N2]eq = 0.5− 0.125= 0.375moldm−3
5. Write Kc = . . . and plug in the values or
expressions of the Equilibrium
amounts. Make sure to answer the
question.
Kc =[NH3]
2
[H2]3 · [N2]
=(0.25)2
(0.625)3(0.375)= 0.68
93
EQUILIBRIUM Relation between ∆G and Kc
7.6 Relation between ∆G and Kc
The position of equilibrium corresponds to: a maximum value of entropy and aminimum in the value of the Gibbs free energy change.
∆G and Kc are related by (from DB 1):
∆G =−RT lnKc
∆G= free energy change [J mol−1]R =gas constant [8.31 J K−1 mol−1]T = temperature [K]Kc =equilibrium constant depends
lnKc =−∆GRT
e lnKc = e−∆GRT
Kc = e−∆GRT
in order to rewite∆G =−RT lnKc to make Kc thesubject, first lnKc is isolated on one side.
The inverse function of ln is: e to the power.
From this equation we can infer that:– the larger∆G (positive value), the smaller Kc will to be.– the opposite– and since e0 = 1, when∆G = 0 then Kc = 1.
Since∆G and Kc are related, they are both indicators for the position of equilibrium andfor the sponteneity of reactions.
reaction reaction
Kc� 1 (almost) no reaction ∆G� 0 non-spontaneousKc < 1 equilibrium favours reactant side ∆G > 0 non-spontaneousKc ≈ 1 approx. equal [reactant] and [product] ∆G ≈ 0 equilibriumKc > 1 equilibrium favours product side ∆G < 0 spontaneousKc� 1 tends to completion ∆G� 0 spontaneous
94
8ACIDS AND BASES
8.1 Acid and base definitions
Brønsted-Lowry Lewis
Acid H+ donor e−-pair acceptorBase An alkali is a base that
is soluble in waterH+ acceptor e−-pair donor
Amphoteric H+ donor & acceptor e−-pair donor & acceptorAmphiprotic H+ donor & acceptor —
Every Brønsted-Lowry acid & base is also a Lewis acid and base:
O
H
H + H+⊕
O
H
H H
But some Lewis acids and bases are not Brønsted-Lowry acids and bases:
N
H
H
H + B H
H
H
H⊕
N
H
H
B
F
F
F
Conjugate acid/base pair a pair of molecules that differ by a single H+-ion
CH3COOH
conjugateacid
+ NH3
conjugatebase
−−*)−− CH3COO−
conjugatebase
+ NH +4
conjugateacid
H+ H+
So the conjugate base is the species formed after the acid has donated a proton,and the conjugate acid is the species formed after the base has accepted aproton.
95
ACIDS AND BASES Strong vs weak
8.2 Strong vs weak
Strong acid/base completely dissociates into its ion in aqueous solutionA proton in solution canbe written as H+ orH3O+.
Strong acid HCl+H2O−−→Cl– +H3O+100%
Strong base NaOH−−→Na++OH–
Weak acid/base dissociates partially into its ion in aqueous solution
In solution, the [H2O]barely changes due tothe reaction, so weassume it staysconstant.Ka/ Kb incorporates itsvalue.
Weak acid CH3COOH+H2O −*)−−CH3COO– +H3O+≈ 1%
Weak base NH3+H2O −*)−−NH +4 +OH–
Ka =[CH3COO−][H3O+][CH3COOH]����[H2O]
Kb =[NH +
4 ][OH−]
[NH3]����[H2O]
Common acids Common bases
StrongHCl hydrochloric acid
H2SO4 sulfuric acid
HNO3 nitric acid
Weak
Any carboxylic acid is aweak acid, and theconjugate base(carboxylate) a weakbase.
H3PO4 phosphoric acid H2PO –4 dihydrogen phosphate
WeakCH3COOH ethanoic acid CH3COO– ethanoate
H2CO3 carbonic acid HCO –3 hydrogen carbonate
Any amine is a weakbase, and the conjugateacid (ammonium) aweak acid.
HCO –3 hydrogen carbonate CO 2–
3 carbonate
CH3NH +3 methyl ammonium CH3NH2 methane amine
NH +4 ammonium NH3 ammonia
The strong bases occuron the exam as metalsalts, such as NaOH,KOH, . . .and Na2O, K2O, . . .
OH– hydroxideStrongO2– oxide
C2H5O– ethoxide
96
ACIDS AND BASES Strong vs weak 8
Experiments to distinguish strong and weak
The difference between strong and weak is the amount of dissociation into ions. Anequimolar amount (!) of a strong acid will have a larger [H3O+] than the same amount ofa weak acid. The same holds for bases, but then the strong base will have a larger [OH– ].
1. pH measurement – strong acids have higher [H3O+] so the pH will be lower thanthat of a weak acid, and the reverse holds for strong bases. The pH can bemeasured with a digital pH meter, or with a (universal) indicator.
2. Conduction measurement – strong acids/bases dissociate fully into ions, so theconductivity of the solution will be much higher.
3. Reaction rate – strong acids have higher [H3O+], so the reaction rate will behigher. For example: an equimolar amount of strong acid will produce a morevigorous reaction with a reactive metal than a weak acid.
Typical reactions of acids for which observations can be made include:
Reaction type Example reaction Observation
neutralisation 2HCl+Na2O−−→ 2NaCl+H2O exothermic, so the T ↑
metals (redox) 2HCl+Mg−−→MgCl2+H2 H2(g) bubbles
carbonate 2HCl+Na2CO3 −−→ 2NaCl+CO2+H2O CO2(g) bubbleshydrogen carbonate HCl+NaHCO3 −−→NaCl+CO2+H2O CO2(g) bubbles
Important distinctions
Strong fully dissociated into ions
Concentrated solution with a highconcentration
Corrosive highly reactive chemical
Weak partially dissociated to ions
Dilute solution with a lowconcentration
TWO factors that affect the pH/pOH of a solution the strength of theacid or base (strong vs weak) AND the concentration of the acid or base
A highly concentrated solution of a weak acid can easily be more acidic than adilute solution of a strong acid.
97
ACIDS AND BASES pH scale
8.3 pH scale
pH stands for potential of Hydrogen, which is a scale to specify the acidityor basicity of an aqueous solution. The scale is logarithmic, so a changeof one unit in pH represents a 10-fold change in concentration.
Formula to calculate p. . . Inverse to calculate [. . . ]
pH=− log [H3O+] [H3O+]= 10−pH
pOH=− log [OH−] [OH−]= 10−pOH
Neutral solution (at 298 K pH = 7)solution that contains equal amounts of H3O+ and OH–
Acidic solution (at 298 K pH < 7)solution that contains H3O+ (more than OH– )
Alkaline solution (at 298 K pH > 7)solution that contains OH– (more than H3O+)
Contrary to popularbelief, the pH value canbe less than 0 orgreater than 14 forstrong andconcentrated acids andbases.
pH [H3O+] [OH−] pOH
· · · · · · · · · · · ·−1 101 10−15 15
0 100 10−14 14
1 10−1 10−13 13
· · · · · · · · · · · ·6 10−6 10−8 8
7 10−7 10−7 7
8 10−8 10−6 6
· · · · · · · · · · · ·13 10−13 10−1 1
14 10−14 100 0
15 10−15 101 −1
· · · · · · · · · · · ·
acidic
neutral
alkaline
acidic
neutral
alkaline
So: +1 on the pH scale corresponds to a 10× larger [H3O+]
98
ACIDS AND BASES pH scale 8
pH calculations: strong acids and bases
Calculate the pH of a 0.500 mol dm−3 Na2O-solution
1. Write the r.eq. of acid OR base with H2O Na2O+H2O−→ 2Na++ 2OH –
2. Use molar ratios to determine the
[H3O+] or [OH – ]
The molar ratio of Na2O : OH – = 1 : 2.
So: [OH – ]= 2 · 0.500= 1.00moldm−3
3. Answer the question
(by using the pH formulas)
pOH=− log [OH – ]
=− log1.00moldm−3 = 0.00pH= 14.00− 0.00= 14.00
pH calculations: weak acids and bases
Calculate the pH of a 0.500 mol dm−3 CH3COOH-solution
1. Write the r.eq. of acid OR base with H2O CH3COOH+H2O−*)− CH3COO – +H3O+
2. Write Ka = . . . or Kb = . . .Ka =
[CH3COO−][H3O+][CH3COOH]
3. Equate the expression to:
K=x2
M − xwhere x = [H3O+] or [OH – ] and
M = the initial [acid] or [base]
Ka =[CH3COO−][H3O+][CH3COOH]
=x2
M − x
4. Of three variables, two must be given:
1. Ka or Kb from values in the DB
2. the initial [acid]0 or [base]0
3. x = [H3O+] or [OH – ], can be
calculated from the pH
From DB: pKa = 4.76 for CH3COOH
Ka = 10−pKa = 10−4.76
M = 0.500moldm−3
5. Plug the two known values into the
expression and use the GDC intersect
function to calculate the unknown
Ka =x2
M − x
10−4.76 =x2
0.5− xGDC intersect gives:
x = 2.9× 10−3 moldm−3 = [H3O+]
6. Answer the question
(by using the pH formulas)
pH=− log [H3O+]
=− log2.9× 10−3 moldm−3
= 2.53
99
ACIDS AND BASES Buffers
8.4 Buffers
Buffer a buffer solution resists changes in pHwhen relatively small amountsof acid or base are added
Buffer composition a solution containing a weak acid with its conjugateweak base in approximately equal amounts
pH of a buffer approximately equal to the pKaof the weak acid
Acidic buffers Alkaline buffersall (halogenated) carboxylic acids all amines incl. NH3
To prepare a solution with the buffer composition we can mix:1. a weak acid + its conjugate weak base (approx. equal amounts)2. a weak acid in excess + strong base3. a strong acid + weak base in excess
When a strong acid or base is added to a weak base or acid, first a completion reactionoccurs. When this reaction completes, the resulting solution will have the buffercomposition: weak acid + its conjugate weak base.
For example, a buffer solution forms when 2.0 mol HA (excess weak acid) is mixed with asolution containing 1.0 mol NaOH (limiting strong base). The result of the completionreaction is a solution with equal amounts of weak acid + conjugate weak base (= buffer):
HA + NaOH −−→ A– + Na+ + H2O
before reaction 2.0 mol 1.0 mol 0 mol – –after reaction 1.0 mol 0 mol 1.0 mol – –
8.5 pH curves
Titration analytical method to accurately determine the concentration of asubstance (analyte), by reference to a known standard solution (titrant)
Analyte substance under investigation, unknown concentration
Titrant substance that reacts with the analyte, with known concentration
Equivalence point (EP) where the amount of analyte equals the amount oftitrant. Indicators show when the equivalence point is reached.
100
ACIDS AND BASES pH curves 8
In acid/base titrations, the concentration of an acid in solution can be determined, byreference to a known base solution, and vice versa.
The acid/base reaction that occurs must be a completion reaction, otherwise theequivalence point cannot be accurately determined. This means that one at least onecomponent in the titration has to be strong, or both. How the pH changes during atitration can be visualised in pH curves.
There are two possible situations: 1. both the analyte and titrant are strong, or 2. one isstrong and the other weak. Below, two examples are given for both situations.
1. strong acid + strong base 2. weak acid + strong base
Analyte HCl CH3COOH
Titrant NaOH NaOH
Reaction HCl+NaOH−−→NaCl+H2O CH3COOH+NaOH−−→NaCH3COO+H2O
pH curves pH
cm3 NaOH sol.
1
7
13
EP
VEP
indicatorrange
pH
cm3 NaOH sol.
1
7
13
VEP
EP when amount HCl = amount NaOH, when amount CH3COOH = amount NaOH,in other words when [H3O+] = [OH−], during titration NaCH3COO forms,which is neutral so pH= 7 at EP which is alkaline so pH> 7 at EP
Indicator bromophenol blue phenolphthalein
Characteristicfeatures
At 12VEP, the amount of NaOHadded =
12CH3COOH present. At this point half theweak acid is converted into its conjugate base andforms a buffer. At 1
2VEP, pH= pKa
101
ACIDS AND BASES Acid Deposition
8.6 Acid Deposition
Acid deposition acidic particles leave the atmosphere. 2 types: wet (acidrain) and dry (gaseous deposition)
Regular rainwater naturally acidic (pH= 5.6) due to the presence ofdissolved CO2
Acid rain made more acidic by SOx and NOx (fog, dew, snow, rain)
Sources of SOx– Volcanoes– Combustion of S containing fossil fuels
(coal, diesel, . . . ):
S+O2 −−→ SO2
Formation of SO3
2SO2+O2 −−*)−− SO3
Wet deposition
SO2+H2O−−*)−−H2SO3 (weak acid)SO3+H2O−−*)−−H2SO4 (strong acid)
Sources of NOx– Electrical storms & bacteria– Any combustion engine
(air at high temperature and pressure):
N2+O2 −−*)−− 2NO
Formation of NO2
2NO+O2 −−*)−− 2NO2
Wet deposition
2NO2+O2 −−→ HNO2 + HNO3weak strong
Environmental effects
Acid rain triggers a number of inorganic and biochemical reactions with detrimentalenvironmental effects. To counteract its effects we should 1. Switch to alternativemethods of energy production (not fossil fuels) and use less energy (e.g., public transport,reduce consumption). And 2. reduce SOx and NOx production by cleaning exhaustgases using catalytic converters and removing S before, during and after combustion(scrubbing).
Vegetation The soil quality degrades because: 1. nutrients (Mg2+, Ca2+, . . . ) areremoved leading to stunted growth in plants and 2. poisonous Al3+ ions aredischarged by chemical erosion of rocks, which damages the roots preventing theplants to take up water.
Lakes & rivers Acids damage mucous membranes, aquatic life (fish, snails, insect larvae)is very sensitive to pH. Lime CaO / Ca(OH)2 is sometimes added to lakes toneutralise acidity. Additionally, the poisonous Al3+ ions that is discharged bychemical erosion of rocks is poisonous to fish.
Human health Acids also damage human mucous membranes, causing respiratoryilnesses such as asthma, bronchitis, . . .
102
ACIDS AND BASES Acid Deposition 8
Buildings & structures Marble and limestone consists of the insoluble CaCO3, and thecarbonate is a base which reacts with acid to form CO2 and soluble compounds:CaCO3+H3O+ −−→Ca2++CO2+H2O.
103
9REDOX
Oxidation is loss of electrons, the reducing agent loses electrons.
Mg−−→Mg2++ 2 e− Mg is oxidized (loss of e−),the oxidation state increases.
Reduction is gain of electrons, the oxidising agent gains electrons.
O2+ 4 e− −−→ 2O2– O2 is reduced (gain of e−),the oxidation state decreases.
Since electrons are transferred in a redox reaction, you might think that thecharges change. This is not always the case, instead the oxidation state changes:
Oxidation state hypothetical charge that an atom would have, if all bondsbetween different elements were 100% ionic (no covalent component).
Note that the oxidation state of, for example Mg2+, is written as +2, while thecharge is written as 2+.
Redox reaction reaction between an oxidising and a reducing agent,characterised by the transfer of electrons. In all redox reactions, theoxidation state of at least one atom changes.
Memorize: OIL RIG
OxidationIsLoss of electrons,
ReductionIsGain of electrons
Students are oftenflustered that the verb‘oxidised’ and thenouns ‘reducing agent’/ ‘reduced species’belong together.
Just remember that OILRIG applies to the verband the opposites areused for the nouns‘agent’ / ‘species’.
9.1 Oxidation states
1
2
3
4
5
6
Determine the oxidation state per element (!)
Elements not combined with other elements have oxidation number of 0.(e.g. Fe, Cu, H2, O2, P4, S8, . . . )
F when combined always has oxidation state of −1.
O when combined has oxidation state of −2 except in peroxides.(e.g. H2O2. . .when it is −1)
H when combined has oxidation state of +1 except in metal hydrides.(e.g. LiH, NaH. . .when it is −1)
The oxidation state of ions in an ionic compound are equal to their charge.
The sum of all the oxidation states of a species equals the charge.
105
REDOX Oxidation states
What is the oxidation number of. . .
. . . Cl in ClO –4 ?
Cl= ?4× O =−8 +total = 0
So the ox. state of Cl is +7.
. . . C in H2C2O4?
2×H=+22× C = ?4×O=−8 +total = 0
So the ox. state of C is +3,since the two C atoms have toaccount for +6.
. . . P in NaH2PO3?
Na=+12× H =+2
P = ?3× O =−6 +
total = 0
So the ox. state of P is +3.
Exam
ple.
Identify which species are oxidised.
From the following reaction, deduce whether Zn is oxidised or reduced:
Zn+CuO−→Cu+ZnO
1. Is the element losing or gaining
electrons?
ZnO is an ionic compound with zinc
present as Zn2+. The Zn metal loses two
electrons to form the Zn2+ ion.
2. Apply ‘OIL RIG’ Oxidation is loss, so Zn metal is oxidised.
Identify which species is the oxidising or reducing agent.
From the following reaction, deduce whether Cu2+ is the oxidising or reducing agent:
Cu2++Mg−→Cu+Mg2+
1. Is the element losing or gaining
electrons?
The Cu2+ ion gains two electrons to form
the Cu metal.
2. Apply ‘OIL RIG’ Reduction is gain, so Cu2+ is reduced.
3. Reducing agents are oxidised, and
oxidising agents are reduced
Cu2+ is an oxidising agent.
106
REDOX Reactions 9
9.2 Reactions
The overall balanced redox reaction can be derived from the oxidation andreduction half-reactions. A list of half-reactions can be found in databooktable 24: “Standard electrode potentials at 298 K”.
Half-reactions are used to separate the oxidation and reduction parts of aredox reaction, useful as a tool to balance redox reactions.
Balance redox reactions from the half-reactions.
An acidified potassium permanganate solution reacts with a copper coin. Write down
the balanced redox reaction from the half-reactions.
1. Find both half-equations
(DB 24).MnO −
4 + 8H++ 5e− −→Mn2++ 4H2O | × 2
Cu−→Cu2++ 2e− | × 5
2. Balance e− by multiplying
both half-reactions.2MnO −
4 + 16H++ 10e− −→ 5Mn2++ 20H2O
5Cu−→ 5Cu2++ 10e−
3. Add both half-reactions. 2MnO −4 + 16H++ 10e−+ 5Cu−→
2Mn2++ 8H2O+ 5Cu2++ 10e−
4. Cross out the same
particles on both sides.2MnO −
4 + 16H++���10e−+ 5Cu−→
2Mn2++ 8H2O+ 5Cu2++���10e−
Create a half-reaction when reactant and product are known.
Complete the following half-reaction: N2H4 −→ NO –3
1. Balance elements other than O and H N2H4 −→ 2NO –3
2. Balance O by adding H2O N2H4+ 6H2O−→ 2NO –3
3. Balance H by adding H+ N2H4+ 6H2O−→ 2NO –3 + 12H+
4. Balance charge by adding e− N2H4+6H2O−→ 2NO –3 +12H++10e−
107
REDOX Reactivity
9.3 Reactivity
Standard conditions (−◦ ) all dissolved particles have [. . . ] = 1.00moldm−3,p = 100kPa and T = 298K.
Standard hydrogen electrode (SHE)is the (arbitrary) referencepoint with which theelectrode potentials of otherhalf-reactions are measuredand compared. Thestandard electrode potentialof the SHE is assigned avalue of 0 V under standardconditions.
1.0 mol dm−3 H+
Pt electrode
H2 @ STP
Standard electrode potential E−◦ ox or E−◦ red the potential of a half-reactionunder standard conditions, measured against the SHE. It is a measure ofthe tendency of a substance to oxidise or reduce.
In DB24, all reactions are written in terms of reduction; so the greater the E−◦ value, thegreater the ability to undergo reduction (and be an oxidizing agent). The E−◦ values fromDB24 are only valid under standard conditions. When the conditions change, so do theE−◦ values.
For example, while under standard conditions H2O (+1.23 V) is a stronger reducedspecies than Cl– (+1.34 V). But when the Cl– solution is concentrated, it becomes astronger reduced species than H2O.
Standard electrode potential difference E−◦ difference between E−◦ox and E−◦red
E−◦ = E−◦ (reduction) −E−◦ (oxidation)= E−◦ (oxidised species) −E−◦ (reduced species)
When we calculate E−◦ ,we always doE−◦red− E−◦ox. This chapterisn’t called red−ox fornothing ;-)
By calculating E−◦ , we can determine if a reaction is spontaneous (from thesign) and the voltage of an electrochemical cell.
A reaction is spontaneous when∆G−◦ is negative. From∆G−◦ =−nF E−◦ (DB), we canderive that∆G−◦ is negative when E−◦ is positive, due to the − sign in the formula.
108
REDOX Reactivity 9
We can also easily deduce if a reaction is spontaneous from the databook: when thereduced species is above the oxidized species the E−◦ is positive and∆G−◦ is negative.
Gibbs free energy and standard electrode potential difference
The standard Gibbs free energy can be calculated from E−◦ :
∆G−◦ =−nF E−◦
∆G−◦ =Gibbs free energy [kJ mol−1]n =number of e– transferred [mol]F =Faraday’s constant [9.65× 104 C mol−1]E−◦ =potential difference [V]
Activity series of metals
The activity series ranks metals according to their reductionpotential. Noble metals resist oxidation, while base metalsreadily oxidise. (A base-matal is the opposite of a noble metal,it is not a base). So Au resists oxidation, while Li promotesoxidation.
The complete activity series is shown in DB 25. Note theposition of H: metals below H (such as Cu) do not oxidise inan acidic solution.
Li metal will react with other metal cations, since it is such astrong reducing agent. The more (re)active metal will donatee− to the lesser active metal cation.
LiNaMg
C
ZnFePb
H
CuAgAu
increasing activity
Determine the order of activity from a set of reactions.
What is the correct order of reactivity of the metals X, Y and Z based on the following
equations?
1. XCl+Y−→YCl+X2. ZCl+X−→XCl+Z
1. The more (re)active metal donates e−. 1. Y donates electrons, so Y > X
2. X donates electrons, so X > Z
2. Combine the activities in an ordered list Y is more reactive than X is more reactive
than Z
109
REDOX Electrochemical cells
9.4 Electrochemical cells
Voltaic cell converts chemical energy from spontaneous chemical reactionsto electrical energy.
electrode
Cu
electrode
Zn
salt bridge
anions
cations
Cu2+
Zn2+
half-cell I half-cell II
Cu2++ 2e− −−→Cu Zn−−→ Zn2++ 2e−
electrolytesolution
voltmetere−e−
Half-cell consists of an electrode in contact with an electrolyte: an aqueoussolution which contains ions. A half-cell physically separates the red. &ox. half-reactions, and the electrons flow through the external circuitwith a certain potential difference (measured in volts).
Cell diagram convention a shorthand notation for a voltaic cell, showingthe substances of the different components.
||electrolyte I|electrode I electrode II | electrolyte II
half-cell I half-cell II
||Cu2+|Cu Zn2+ | ZnExample:
Salt Bridge contains an aqueous solution of ions that move across the saltbridge to neutralize the build up of charge in both half-cells to maintainthe potential difference.
To find out in which direction anions and cations flow through the salt bridge,first determine the direction of the flow of electrons (using An OIL RIG Cat).AN OIL RIG CAT
ANodeOxidationIsLoss of electrons,
ReductionIsGain of electrons
CAThode.
– Cations will go in the same direction as the electrons, to neutralize thecharge. So from anode to cathode through the salt bridge.
– Anions will go in the opposite direction as the electrons, to neutralizethe charge. So from cathode to anode through the salt bridge.
110
REDOX Electrochemical cells 9
Electrolytic cell converts electrical energy to chemical energy, bringingabout non-spontaneous chemical reactions.
Since the reaction isnon-spontaneous, theelectrolytic cell does notneed to be physicallyseparated into half-cells,but it may.An electrolytic cell needs apower source, indicatedwith two lines: the longeris the positive terminal.
e− e−+ −
power source
electrodePt
electrodePt
Na+ Cl–
electrolyte solution
Examples of chemically inert electrodes (conductive substances that do not react) arePt & C. They can be used in both voltaic and electrolytic cells.
Determine which half-reactions occur in an electrolytic cell.
Determine which half-reactions occur when a dilute NaCl solution is electrolysed.
Find the experimental setup in the definition box above.
1. List which particles are present at the anode (+)
and cathode (−) separately.
Anode (+): Pt, H2O, Cl –
Cathode (−): Pt, H2O, Na+
2. Apply AN OIL RIG CAT, find the strongest reduced
and oxidised species and copy the half-reactions:
Anode (+): ox. half-reaction (reduced species)
Cathode (−): red. half-reaction (oxidised
species)
Anode (+): strongest oxidation
H2O−→ 12O2+ 2H++ 2e –
Cathode (−): strongest
reduction
H2O+ e – −→ 12H2+OH –
An important deduction from the above example can be made: O2 is formed when waterreacts at the anode, while H2 is formed when water reacts at the cathode. You canmemorize this by an extension of AN OIL. . . So O2 forms at the anode!
Note that it matters whether dilute, concentrated or molten NaCl is electrolysed. Thestrongest oxidation half-reaction in each of these three cases is:
– In dilute NaCl: H2O−−→ 12O2+ 2H++ 2e– (just like DB.24 shows).
– In concentrated NaCl: Cl– −−→ 12Cl2+ e– instead of H2O.
– And while H2O is present in both dilute and concentrated solutions, it is not whenan ionic compound is in its liquid state (molten). The strongest oxidation andreduction half-reactions are: Cl– −−→ 1
2Cl2+ e– and Na++ e– −−→Na.
111
REDOX Electrochemical cells
Voltaic cell and electrolytic cells compared
AN OIL RIG CAT reminds us that oxidation occurs at the anode, and reduction occurs atthe cathode. But while oxidation always occurs at the anode, whether the electrode is+ or− depends on whether we are dealing with a voltaic cell or an electrolytic cell. Takingthe voltaic cell as the standard cell, or the one you learned about first, the anode is − justlike anions have negative charge. In the electrolytic cell the reverse holds.
voltaic cell electrolytic cell
ANode Oxidation occurs here − electrode + electrodeCAThode Reduction occurs here + electrode − electrode
In a voltaic cell: chemical energy is converted to electrical energy. The reaction isspontaneous, E−◦ is positive and the value = the voltage of the voltaic cell (battery).
In an electrolytic cell: electrical energy is converted to chemical energy. The reaction isnot spontaneous, E−◦ is negative and the value = the minimum voltage of thepower source that is required to start the reaction.
Observations and the amount of products formed
When chemical reactions occur, the amounts of reactants are reduced and the amounts ofproducts are increased. The change in the abundance of chemicals can be linked toobservations:
Colour: the colour intensity decreases if a reactant is coloured, and the color intensityincreases if a product is coloured. Remember that transition metal ions are oftencoloured, in particular Cu2+ is blue.
Solid deposition: a solid can deposit onto an electrode, changing the appearance andincreasing the mass.
Gas discharge in a solution: bubbles form when a gas is formed in solution. Inparticular: O2 can form at the anode and H2 at the cathode, given that H2O reactsat the anode or cathode respectively.
pH: the pH changes when the concentrations of H3O+ or OH– change, which can bemeasured.
The mole amounts of reactants / products in an electrolytic cell depend on three factors:1. Current (amount of electrons that pass through the cell per second)2. Duration (the longer, the more electrons pass through the cell)3. Charge on the ion (the higher, the more electrons required to make a neutral atom)
112
REDOX Electrochemical cells 9
Electroplating
The process of using electrolysis to deposit a metal on top of a conductive object thatmust satisfy three conditions.
1. Electrolyte must contain ions of the metal to be deposited2. Cathode is made of the object that will be plated/coated (conductive)3. Anode can also be made of the metal that will be deposited to refill the ions in the
electrolyte
Calculate the amount that reacts/forms in an electrolytic cell.
Calculate the mass of copper that is deposited on the cathode during the electrolysis of
a CuCl2 solution with a current of 5.0 A for 30 minutes.
1. Write the half-reaction that occurs. Cu2+(aq)+ 2e – −→ Cu(s)
2. Calculate number of moles of e – :
n(e−) =current (A)× time (s)Faraday’s constant (F)
Note: this formula is not in the DB.
n(mol)e− =5.0A× 30min× 60smin−1
9.65× 104
= 9.3× 10−2 mol of electrons
3. Use the mole ratio from the half-reaction
to calculate moles of the desired
substance.
Since the mole ratio Cu(s) : e− = 1 : 2,
n (Cu(s))=9.3× 10−2 mol
2=
4.7× 10−2 mol Cu(s)
4. Convert to the required units. m= n×Mm
m= 4.7× 10−2 mol× 63.55gmol−1
= 3.0g Cu(s)
113
REDOX The Winkler method and the BOD
9.5 The Winkler method and the BOD
Winkler method two sequential titrations that are performed on surfacewater to measure the initial amount of dissolved oxygen, and theamount after some pre-defined time to determine the BOD.
Biochemical Oxygen Demand (BOD) the amount of oxygen used todecompose organic matter in a sample of water over a specified timeperiod.
If there is much organic matter (bad), much oxygen will be used up by bacteriaduring decomposition. And if there is little dissolved oxygen to begin with(bad), aquatic life will struggle to survive.
Thermal Pollution: water used in cooling engines in factories tend to be released into awater system, which decreases the dissolved oxygen content.
Organic Matter: a greater amount of organic matter in water means moremicro-organisms will be present, thus decreasing the dissolved oxygen content.This usually occurs due to eutrophication: the excess use of fertilizers anddetergents increases the amount of ‘food’ for bacteria and algea, increasing theBOD.
The Winkler determination of BOD is based on a sequence of redox reactions. The moleratio between O2(aq) = dissolved oxygen (analyte) : S2O 2–
3 (titrant) = 1:4.
114
10ORGANIC CHEMISTRY
10.1 Fundamentals of organic chemistry
Empirical formula shows the simplest whole number ratio of atoms in a compound.
Molecular formula shows the actual number of atoms in a molecule.
Structural formula shows the bonds between atoms, that form a molecule. It can be shownin full, condensed and skeletal form.
structural formulas molecular empiricalfull condensed skeletal formula formula
H C
H
H
C
H
H
C
H
H
H CH3CH2CH3 C3H8 C3H8
H C
H
H
C
H
OH
OH CH3CH(OH)2
OH
OH
C2H6O2 CH3O
H C
H
H
C
H
C
H
C
H
H
H CH3CHCHCH3 C4H8 CH2
A note on skeletal structures: the C-atoms are not explicitly drawn, but a C-atom with thecorrect number of H-atoms is implied at the start/end of a line, and where lines intersect.
H C
H
H
C
C H
H
H
H
C
H
H
C
H
H
OH is the same as OH
115
ORGANIC CHEMISTRY Fundamentals of organic chemistry
Homologous series succesive members differ from each other by −CH2−.
alkane homologous series alcohol homologous series
name structure BP name structure BP
methane CH4 −162 ◦C methanol CH3OH 64 ◦Cethane CH3CH3 −89 ◦C ethanol CH3CH2OH 78 ◦Cpropane CH3CH2CH3 −42 ◦C propan-1-ol CH3CH2CH2OH 98 ◦Cbutane CH3CH2CH2CH3 0 ◦C butan-1-ol CH3CH2CH2CH2OH 118 ◦Cpentane CH3CH2CH2CH2CH3 36 ◦C pentan-1-ol CH3CH2CH2CH2CH2OH 138 ◦C
Members of a homologous series have similar chemical properties (since they share the same functionalgroup) and they show a gradual change in physical properties (for example their melting-/boiling pointsincrease as a result of increasingly stronger London dispersion forces).
10.1.1 Classification of hydrocarbons
Hydrocarbons are compounds containing only hydrogen and carbon atoms.
Saturated hydrocarbonscontain only C−C (alkane)
H C
H
H
C
H
H
H
Unsaturated hydrocarbonscontains C−−C (alkene) or C−−−C (alkyne)
C
F
F
C
F
F
H C C H
Aliphatic compoundscompounds without benzene rings
H C
H
H
C
H
C HH
H
C
H
H
H
Aromatic compoundscompounds with at least one benzene ring
NO2
116
ORGANIC CHEMISTRY Fundamentals of organic chemistry 10
Number of C neighbours
H3C CH
CH3
CH2 C
CH3
CH3
CH3
primary 1◦ directly bonded to one C atomsecondary 2◦ directly bonded to two C atomstertiary 3◦ directly bonded to three C atomsquaternary 4◦ directly bonded to four C atoms
Alcohols, halogenoalkanes, amines and carbocations can be primary (1◦), secondary (2◦)or tertiary (3◦). The atom of interest is the C atom that is bonded to −OH or −X or thatcarries the positive charge. For amines it is different, the atom of interest is the N atom.
alcohol halogenoalkane amine carbocation
primary (1◦) CH3 C
H
H
OH CH3 C
H
H
X CH3 N
H
H CH3+C
H
H
secondary (2◦) CH3 C
H
CH3
OH CH3 C
H
CH3
X CH3 N
CH3
H CH3+C
CH3
H
tertiary (3◦) CH3 C
CH3
CH3
OH CH3 C
CH3
CH3
X CH3 N
CH3
CH3 CH3+C
CH3
CH3
10.1.2 Functional groups
Chemical class a family of compounds that share the same functional group.
Functional group gives distinct chemical properties to a compounds, thereactive part of the molecule.
Sometimes the class ofcompounds has adifferent name as thefunctional group. Thefunctional group nameshave come up quiteoften on IB-exams, besure you know theirnames.
chemicalclass
functionalgroup
contains root examples
alkane alkyl C−C only alkane CH3 CH2 CH3
alkene alkenyl C−−C alkene CH2 CH CH3
alkyne alkynyl C−−−C alkyne H C C H
arene�
benzenederivative
�phenyl
R CH3
117
ORGANIC CHEMISTRY Fundamentals of organic chemistry
chemicalclass
functionalgroup
condensed full structural suffix/prefix examples
carboxylicacid
carboxylCOOH C
O
OH
-oic acidCH3 C
O
OH
aldehyde aldehyde −CHOC
O
H
-alC
O
H5C6 H
ketone carbonyl −CO−C
O
C C
-oneH3C C
O
CH3
alcohol hydroxyl −OH -ol CH3 OH
amine amine −N(H,R)2 -amine H2N CH3
nitrile nitrile −CN C N N C CH3
ether alkoxy −O− C O C alkoxy- CH3 O CH3
halogen −X halogeno- CH3 Br
side-chain−CH3,−CH2CH3. . .
alkyl- CH3 CH
CH3
CH3
ester ester −COO−C
O
O CCH3 CH2 O C
O
CH3
amide carboxamide −CON(H,R)2C
O
N
H
H N
H
C
O
CH3
118
ORGANIC CHEMISTRY Fundamentals of organic chemistry 10
10.1.3 Nomenclature
Naming organic molecules
Determine the name of the following compound: C
H
H
C
CH2OH
CH2 CH3
1. Determine the longest uninterrupted chain,
including the functional group(s)
alk: meth-, eth-, prop-, but-, pent-, hex-
types: -ane, -ene, -yne
H
C
H
C
CH2OH
CH2 CH3
root: propene
the longest chain with
the −OH and C−−C in-
cluded is three, not four.
2. Determine the functional groups
suffix: -oic acid, -al, -one, -ol, -amine
prefix: halogeno-, alkoxy-, alkyl-
multiple: di-, tri-, tetra-. . . etc
H
C
H
C
CH2OH
CH2 CH3
suffix: -ol
prefix: -ethyl
3. Number as low lowest possible:
1 C-atom with the functional group
2 double and triple bonds
3 other
H
C3
H
C2
C1H2OH
CH2 CH3
start counting so that
the functional group
−OH is lowest.
4. Write full name: prefixrootsuffix
1 dash ( - ) between numbers and letters
2 comma ( , ) between numbers
H
C
H
C
CH2OH
CH2 CH3
2-ethylprop-2-ene-1-ol
Cl C
Cl
Cl
C
O
OH
C
O
H
C C C
H
O
C
H
H
C
H
O CH2 CH2 CH3
1. root: ethane root: butyne root: ethene
2. suffix: -oic acid
prefix: trichloro-
suffix: -dial
prefix: —
suffix: —
prefix: propoxy-
3. Cl C2
Cl
Cl
C1
O
OH
C1
O
H
C2
C3
C4
H
O
C2
H
H
C1
H
O CH2 CH2 CH3
4. 2,2,2-trichloroethanoic acid but-2-yne-1,4-dial 1-propoxyethene
119
ORGANIC CHEMISTRY Fundamentals of organic chemistry
10.1.4 Resonance structure of benzeneThe simplest aromatic hydrocarbon is benzene, C6H6.
resonance structures
or
resonance hybrid
The p orbitals on thesp2 hybridized C atomsoverlap sideways above& below the molecule,resulting in a ring ofdelocalised π electrons.
The six C C are each equally strong and long (bond order is 1.5), instead of havingthree C C and three C C bonds (of bond orders 1 and 2). The strength and lengthof the C C bonds are somewhere between that of C C and C C.
Chemical proof: while regular alkenesreadily undergo addition reactions,benzene resists addition and insteadundergoes substitution reactions.
+ Br2
Br
+ HBr
Physical proof: the following table showsmeasured bond strengths and lengths.
bond strength bond lengthin kJ mol−1 in nm
C C 346 0.154C C 614 0.134C C 507 0.140
120
ORGANIC CHEMISTRY Isomers 10
10.2 Isomers
Isomers are molecules that have something in common, but are different.
Stereoisomers have different spatial arrangements (→ both cis/trans and optical isomers).
Chiral atom contains 4 different groups→ optical isomers.
structural isomers cis/trans and E/Z optical isomers
same molecular formula, butdifferent structural formula
different spatial arrangement ofatoms around C−−C or ring due tolack of free rotation
different spatial arrangement offour different groups on a chiralC-atom
C
R1
R2
C
R3
R4
∗C R2
R1
R4
R3
requirement
R1 6=R2 and R3 6=R4 R1 6=R2 6=R3 6=R4
examples
C4H10 cis/trans chiral atom/centre
H3C CH3 CH3 CH3
H3C CH
CH3
CH3C
F
H
C
F
H
cis: same
C
F
H
C
H
F
trans: opposite
C
H
H
H
C
H
H
∗C
Cl
H
C
H
H
H
C2H6O E/Z drawing isomers
H3C CH2 OH
H3C O CH3
C
H
CH3
C
F
Cl
Z: same
C
H
CH3
C
Cl
F
E: opposite
∗C
C2H5
ClCH3
H
∗C
C2H5
ClH3C
H
mirror
chemical & physical properties
(very) different/distinct and can(easily) be separated
similar, but dipole moment may bedifferent
all chemical and physicalproperties are the same; except therotation of plane polarized lightand reactions with enzymes
121
ORGANIC CHEMISTRY Isomers
10.2.1 Structural isomers
Structural isomers same molecular formula, different structural formulas
There are three sub-categories of structural isomers, that correspond to threestrategies when drawing structural isomers:
Generally speaking,branched isomers existwhen molecules have 4or more C-atoms.1. chain isomers: arise because of the possibility of branching in carbon chains.
C4H10
CH3 CH2 CH2 CH3
CH3 CH
CH3
CH3
C5H12
CH3
CH2
CH2
CH2
CH3
CH3 CH
CH3
CH2 CH3
CH3 C
CH3
CH3
CH3
2. positional isomers: the carbon skeleton remains unchanged, but groups have different positions.
C3H6X2
CH
X
X
CH2 CH3CH2
X
CH
X
CH3
CH2
X
CH2 CH2
X
CH3 C
X
X
CH3
3. functional group isomers: the isomers contain different functional groups.different positions.
C2H6O
CH3 CH2 OH
CH3 O CH3
alcohol
ether
C3H6O
CH3 C
O
CH3
CH3 CH2 C
O
H
ketone
aldehyde
C2H4O2
CH3 O C
O
H
CH3 C
O
OH
ester
carboxylic acid
C3H6
CH2 CH CH3alkene
cyclic
122
ORGANIC CHEMISTRY Isomers 10
When you are asked to draw the structural isomers from a given molecular formula,think about all three types. Ask yourself: Can you branch the carbon chain? Can youmove a group around on that chain? Is it possible to change the functional group?
Finding structural isomers
Find all the structural isomers that exist of C4H10O.
1. Branches: Is
branching a
possibility?
C C C C
C C
C
C
2. Positions: Draw any one
isomer. If any, can the
position of the functional
group be drawn at a different
position?
CH2
OH
CH2 CH2 CH3
CH3 CH
OH
CH2 CH3
CH3 C
OH
CH3
CH3
CH2
OH
CH
CH3
CH3
3. Functional group: Can you
draw another functional group
instead of the previous? Also
change the position of this group.
CH3 CH2 O CH2 CH3
CH3 O CH2 CH2 CH3
CH3 O CH
CH3
CH3
123
ORGANIC CHEMISTRY Isomers
10.2.2 cis/trans and E/Z isomers
geometric isomers synonym for cis/trans and E/Z.This occurs when rotation in a molecule isrestricted due to a double bond (or a ring).
There are no geometric isomers if R1 =R2 or R3 =R4.
C
R1
R2
C
R3
R4
cis/trans isomers when two groups are identical, those can be on the sameside (cis) or on the opposite side (trans). Otherwise→ E/Z.
CIP priority rule atoms with larger molar mass receive higher priority.1. Compare the mass of the first atom.2. If they are the same, compare the masses of the next atom etc.
E/Z isomers when all the four R-groups are all different, use CIP todetermine whether the groups with highest priority are on the zameside (Zusammen = together) or on the opposite side (Entgegen =opposite)
cis/trans isomers
Since internal rotation is limited around a double bond, molecules that have groups onthe same side (cis) of the double bond are different from molecules with those groups onopposite sides (trans) of the double bond.
H C
Cl
H
C
Cl
H
H H C
Cl
H
C
Cl
H
H
molecules are the same since the C−Cbond allows rotation.
C
Cl
H
C
Cl
H
“cis”
C
Cl
H
C
Cl
H
“trans”molecules are not the same since the C−−C
bond does not allow free rotation.
C
H
H
C
CH3
Clno cis/trans isomers since the left C has
two identical groups.
C
H
F
C
CH3
Clcannot assign cis or trans since no two
groups are the same→ E/Z
124
ORGANIC CHEMISTRY Isomers 10
E/Z isomers
C
CH3
H
C
CH3
Cl
trans-2-chloro-2-buteneZ-2-chloro-2-butene
C
CH3
H
C
CH3
Cl
cis-2-chloro-2-buteneE-2-chloro-2-butene
It appears that cis (on the same side) andZusammen (together) mean the samething, but they do not! The rules todetermine which is which differ; for
cis/trans we look at the positions of thesame groups over the double bond and for
E/Z we look at the positions of thegroups with highest CIP priority over the
double bond.
To determine the name using the E/Z system, look at each C in the C−−C separately andassign a priority (1 or 2) to both groups attached to each of the C’s. In the abovemolecule we should assign priority as follows:
10.2.3 Optical isomers
→ follow updates on learn.ib.academy
125
ORGANIC CHEMISTRY Reactions
10.3 Reactions
10.3.1 CombustionCombustion is an exothermic reaction between a fuel and oxygen. Based on theelemental composition of a fuel, we can predict which products form after combustion.
element in fuel combustion product averse effects
H H2O —
C CO2 (complete) greenhouse gas
CO (incomplete) poisonous, irreversibly binds to haemoglobin
C (incomplete) irritation to the repository tract
S SO2 acid deposition
N NO acid deposition
Dependent if oxygen isin excess or limited,complete combustion(→ CO2) or incompletecombustion (→ COand/or C) occurs.
10.3.2 Substitution
Free radical substitution reaction a halogen atom (−X) replaces a hydrogenatom (−H) in alkanes, under the influence of light.
H C
H
H
H + Br BrUV-light
H C
H
H
Br + H Br
Alkanes have strong non-polar C−C and C−H bonds, hence alkanes arerelatively inert (unreactive). Under the influence of (UV-)light, halogens (X2)can form very reactive radicals that can replace H-atoms in alkanes.
Nucleophilic substitution reaction a ‘stronger’ nucleophile such as OH–
replaces a ‘weaker’ nucleophile such as −X
H C
H
H
Br + OH− H C
H
H
OH + Br−
Halogenoalkanes are more reactive than alkanes since they possess polar bonds+δC
−δX. The+δ
C-atom can be ‘attacked’ by electron-rich nucleophiles.Moreover, C−X bonds are typically weaker than C−H bonds making −Xeasier to replace. Since halogenoalkanes are more reactive than alkanes, nolight is required to initiate the reaction.
126
ORGANIC CHEMISTRY Reactions 10
10.3.3 Addition
Addition reaction the π bond in alkenes or alkynes opens, and new bondsform to each of the two carbons around the original π bond.
CH
H
C H
H
+ P Q CH
H
P
C H
Q
H
π bonds are somewhat weaker than σ bonds, so alkenes and alkynes are morereactive than alkanes resulting in the fact that no light is needed for reaction.Alkenes and alkynes will undergo addition reactions instead of substitution.
addition of example
hydrogenation→ alkane
H H CH
H
C H
H
+ H HNi
heatCH
H
H
C H
H
H
halogenation→ dihalogeno
X X CH
H
C H
H
+ Br Br CH
H
Br
C H
Br
H
halogenation→ halogeno
H X CH
H
C H
H
+ H Cl CH
H
H
C H
Cl
H
hydration→ alcohol
H OH CH
H
C H
H
+ H OHH3PO4
heatCH
H
H
C H
OH
H
Differentiate experimentally between alkanes and alkenes/alkynes
Br2 has a distinct brown colour. By adding a drop to an excess alkane or alkene/alkyne,the brown colour completely disappears as Br2 is the limiting reactant:
substitution CH3−CH3+Br2light CH3−CH2Br+HBr
addition CH2−−CH2+Br2 CH2Br−CH2Br
Performing the experiment in the dark however, prevents the substitution reaction fromoccuring: the Br2 would only decolorize if the addition reaction can occur, thusproviding evidence for the presence of double bonds.
127
ORGANIC CHEMISTRY Reactions
Asymmetric addition
When an asymmetric alkene (such as propene CH2−−CH−CH3) reacts with anasymmetric reactant (such as H X or H OH), two structural isomers are formedfrom two different competing reactions.
H C
H
H
C
H
C
H
H + H OH
H C
H
H
C
H
H
C
OH
H
H (propan-1-ol)
H C
H
H
C
OH
H
C
H
H
H (propan-2-ol)
Addition polymerization
Polymer a large molecule that forms by linking many monomers together.Synthetic plastics are an example of polymers.
Addition polymer the monomers react by breaking the double bond,linking on every second C atom.
The trick to drawing polymers is to draw 90° angles in the monomers:
in general:
n C
P
R
C
Q
S
C
P
R
C
Q
S
n
polybut-2-ene:
n C
CH3
H
C
CH3
H
C
CH3
H
C
CH3
H
n
Exam
ple.
The repeating unit of an addition polymer always contains two C atoms, since it reflectsthe structure of the monomer. Reasoning backwards starting from the polymer, it isbetween these two C atoms that the monomer had the double bond.
128
ORGANIC CHEMISTRY Reactions 10
10.3.4 Condensation / esterification reaction
Esterification carboxylic acid + alcohol H2SO4
heatester + H2O
H C
H
H
C
O
OH
ethanoic acid
+ HO C H
H
H
methanol
H C
H
H
C
O
O C H
H
H
methylethanoate
+ H O
H
Naming esters is different from naming other organic compounds. The general name ofesters is alkylalkanoate, where the alkyl part of the name is derived from the alcoholparent and alkanoate is derived from the carboxylic acid parent molecule.
methyl is derived fromthe alcohol, methanol
methyl ethanoateethan signifies twocarbons in the acid,
-oate is the ester suffix
Remember that “alk” in both parts of the name should be replaced by the numeric prefixmeth-, eth-. . . to signify the number of C atoms.
129
ORGANIC CHEMISTRY Reaction mechanisms
10.4 Reaction mechanisms
10.4.1 Definitions and understandings
Reaction mechanism reactions take place gradually or in steps; breakingbonds and forming new bonds. Reaction mechanisms show themovement of single electons with ‘fishhook’ arrows ( ) or themovement of electron pairs with regular arrows ( ) for each step.
Homolytic fission the process of breaking a covalent bond where each of thefragments retain one of the bonding electrons, forming radicals.
A B A + B
Free radical contains an unpaired electron indicated by a dot, a reactivespecies formed when a molecule undergoes homolytic fission.
Cl Br O H C HH
H
Heterolytic fission the process of breaking a covalent bond where onefragment takes both bonding electrons from the other.
A B⊕
A +
B
A nucleophile is a Lewis base, an electrophile is a lewis acid.
Nucleophile electron rich species capable of donating an electron pair toform a covalent bond. The donated electon pair is usually a lone pair,but it can also be the π electron pair in a double or triple bond.
Electrophile electron deficient species capable of accepting an electron pairto form a covalent bond. Usually an ion with a + charge, or any atomwith a δ+ charge.
Notes on drawing arrows the base of an arrow always starts from anelectron pair (lone or bonding pair) and always points to an atom. Also,arrows never pass a reaction arrow.
Nu +
⊕E Nu E
130
ORGANIC CHEMISTRY Reaction mechanisms 10
Steric hindrance is the obstruction in 3d space by bulky groups, preventingsuccessful collision or making the successful collision between reactantsless likely, thus slowing down reactions.
The following table shows that approach from the left becomes increasinglymore difficult due to the greater number of bulkier groups. −H atoms aresmall so they are drawn as a small circle, while alkyl groups such as −CH3 arebulkier and will be drawn as a larger circle.
methane1◦ halogeno-
alkane2◦ halogeno-
alkane3◦ halogeno-
alkane
C X C X C X C X
least sterichindrance
most sterichindrance
Positive inductive effect alkyl groups (such as −CH3, −C2H5. . . ) donateelectron density, thereby reducing the positive charge on thecarbocation, thus stabalising a carbocation. The more alkyl groups thatare attached to a carbocation, the more electron density the carbocationreceives and so the more stable it becomes.
Generally the more stable a compound is, the easier it is to form. So when acarbocation is more stable, the activation energy goes down and with it the rateof its formation goes up.
1◦ carbocation 2◦ carbocation 3◦ carbocation
CH3
+C
H
H CH3
+C
CH3
H CH3
+C
CH3
CH3
least stablecarbocation
most stablecarbocation
The arrows show thepushing of electrondensity by alkyl groupsto the carbocation, notdative/coordinatecovalent bonds.
Carbocation species are highly reactive electrophilic intermediates (C reallyprefers having 4 bonds instead of 3): the formation of carbocationintermediates is always rate determining, the subsequent reaction step willalways rapidly form a covalent bond from a lone-pair on a nucleophile to thecarbocation.
131
ORGANIC CHEMISTRY Reaction mechanisms
10.4.2 Free radical substitution
Free radical substitution reaction a halogen atom (−X) replaces a hydrogenatom (−H) in alkanes, under the influence of light. The mechanismproceeds in three stages: initiation, propagation and termination.
H C
H
H
H + Cl Cl H C
H
H
Cl + H Cl
The free radical substitution reaction of CH4 with Cl2 (above) is depicted below, on theleft the reaction in full structural formulas and the identical reactions on the right incondensed structural formulas:
1. Initiation: produces two radicals from X−X
Cl Cllight
Cl + ClCl2light
2Cl
2. Propagation: the number of radicals in the reaction mixture does not change.
H C
H
H
H + Cl H C
H
H
+ H Cl CH4 + Cl CH3 + HCl
H C
H
H
+ Cl Cl H C
H
H
Cl + Cl CH3 + Cl2 CH3Cl + Cl
3. Termination: two radicals combine to form new molecules.
H C
H
H
+ Cl H C
H
H
Cl CH3 + Cl CH3Cl
Cl + Cl Cl Cl 2Cl Cl2
132
ORGANIC CHEMISTRY Reaction mechanisms 10
10.4.3 Nucleophilic substitution
Nucleophilic substitution reaction a ‘stronger’ nucleophile such as OH–
replaces a ‘weaker’ nucleophile such as −X
R C
R
R
Br + OH− R C
R
R
OH + Br−
SN1 mechanism
C
R
Br
RR
rds C+
R
R Rcarbocationintermediate
OH
+
Br C
R
OH
RR
or
racemic mixture
C
R
RR
OH
SN2 mechanism
H
O ++δ
C
R
RR
Br rds
C Br
R R
OH
R
C
R
RR
OH
stereo inversion
+
Br
SN1 favoured SN2 favoured
1◦ — 3◦ 3◦ carbons have greater positive in-ductive effect: the formation of thecarbocation intermediate is easier
1◦ carbons have less steric hindrance,making the approach of the nucleophileeasier
solvent (carbocat)ions are stabilised through sol-vation in aqueous solutions and otherprotic solvents (solvents able to formH-bonds, so containing−OH or−NH).
aprotic solvents not able to formH-bonds. The reactant OH– is lessstabilised, increasing its nucleophilicstrength, favouring SN2.
For nucleophilic substitution reactions on 2◦ carbons both mechanisms will occur andcompete, so only then does the solvent become directing to favour either mechanism.
133
ORGANIC CHEMISTRY Reaction mechanisms
10.4.4 Electrophilic substitution on benzene
Benzene undergoes substitution reactions rather than addition. The typical reaction is ofbenzene with the very strong electrophile NO +
2 derived from nitric acid (HNO3):
H
+ HNO3H2SO4
NO2
+ H2O
HNO3 alone is not reactive enough to react with the resonance stabilised benzenemolecule. But NO +
2 is; formed from nitric acid (HNO3) with sulfuric acid (H2SO4):
HNO3 + H2SO4
+H2NO3 +
−HSO4
+NO2 H2O
H+
Interestingly, in thisparticular reaction thestrong acid HNO3 reactsas a base by acceptingH+ from H2SO4.
The formed H2NO +3 is highly unstable and dissociates, forming the strongly
electrophilic NO +2 that is required to react with benzene:
H
H
++
NO2
H+
NO2
H
H
NO2
++
H
The mechanism is a great example of how catalysts work: while H2SO4 participates inthe reaction mechanism, the H+ that forms recombines with HSO –
4 to regenerateH2SO4. During this reaction the amount of H2SO4 does not change, but it allows theformation of a reactive species (NO +
2 ) that lowers the activation energy.
134
ORGANIC CHEMISTRY Reaction mechanisms 10
10.4.5 Electrophilic Markovnikov additionWhen an asymmetric reactant, such as
+δH−δX or
+δX−δX (interhalogen), reacts with
an asymmetric alkene such as propene CH3−CH−−CH2, two structural isomers areformed from two different competing reactions. By understanding the mechanism, wemay predict which of the isomers are the major product.
Interhalogens molecules in which only halogens are combined. Thesimplest is X X, where X and X are different, such as:
+δI−δBr
+δBr−δ
Cl+δ
Cl−δF
The addition of+δX
−δX occurs via the same mechanism as that of+δH
−δX.
Markivnikov rule the+δH or
+δX gets attached to the carbon withmore hydrogen substituents, and the halide
−δX attaches to thecarbon with more alkyl substituents (more stabilised carbocation).
H C
H
H
C
H
C
H
H ++δH X
H C
H
H
C
H
H
+C
H
H
1◦ carbocationless stable
+
X H C
H
H
C
H
H
C
X
H
H
minor product
H C
H
H+C
H
C
H
H
H
2◦ carbocationmore stable
+
X H C
H
H
C
X
H
C
H
H
H
major product
First the double bond opens to add the electropositive element under formation of thecarbocation and halide-ion, then the halide-ion bonds to the carbocation.
H C
H
C
H
H ++δBr Br H C
Br
H
+C
H
H +
Br H C
Br
H
C
Br
H
H
induced /temporary
dipole
135
ORGANIC CHEMISTRY Reactions overview and retrosynthesis
10.5 Reactions overview and retrosynthesisSubstitution reactions mechanism
alkane + X2UV-light
halogenoalkane + HX free radical
halogenoalkane + X2UV-light
dihalogenoalkane + HX free radical
halogenoalkane + O
H alcohol +
X SN1 and SN2
Substitution reactions
alkene + H2Ni
heatalkane
alkene + X2 dihalogenoalkane
alkene + HX halogenoalkane
alkene + H2O alcohol
n alkene (polymer)n
electrophillic addition
Condensation reactions
carboxylic acid + alcoholH2SO4
heatester + H2O
Benzene reactions
+ HNO3H2SO4
heatNO2 + H2O electrophillic substitution
NO2
HCl (conc.), Sn⊕NH3
OH–
NH2 + H2O
Alcohol oxidation / carboxylic acid / aldehyde / ketone reduction
1◦ alcoholH+/Cr2O 2–
7
NaBH4 or LiAlH4
aldehyde + 2H+ + 2e–H+/Cr2O 2–
7 +H2O
LiAlH4
carboxylic acid + 2H+ + 2e–
2◦ alcoholH+/Cr2O 2–
7
NaBH4 or LiAlH4
ketone + 2H+ + 2e–
136
11MEASUREMENT AND DATA
PROCESSING
Qualitative data refers to all non-numerical information that can beobtained from observations rather than a measurement. Used toidentify: which compound?
Quantitative data refers to numerical information obtained frommeasurements, always associated with systematic and randomerror/uncertainty. Used to identify: how much compound?
Systematic error To decide if an error issystematic or randomask yourself: wouldrepeating theexperiment (partially)remove the error?
refers to flaws in experimental design and machinecalibration that lead to an error, which is deviated in a particulardirection.
Random Error are statistical fluctuations (in either direction) in themeasured data. This can be caused by:
1. Readability or precision limitations of the measuring equipment2. Uncontrollable changes in the surroundings between trials, such
as temperature variations, air currents, . . .
Accuracy refers to how close trials are to a standard/known literature value.→High systematic error leads to low accuracy.
Precision refers to how close trials are to each other.→High random error leads to low precision.
For example, the boiling point of water at sealevel is measured several times. Two differentbrands of digital temperature probes are used.The temperature readings from brand 1 givehave a larger random error than brand 2. Toreduce the effects of random error we canincrease the number of trials.
brand 1 brand 2Trial 1 102.5 ◦C 96.5 ◦CTrial 2 101.5 ◦C 96.7 ◦CTrial 3 97.5 ◦C 96.6 ◦CAverage 100.5 ◦C 96.6 ◦C
Brand 2 suffers from a large systematic error, but it is very precise. To reduce thesystematic error, the temperature probe could be re-calibrated.
137
MEASUREMENT AND DATA PROCESSING Graphical Techniques
11.1 Graphical Techniques
Sketched graphs have labelled but unscaled axes, and are used to showtrends, such as variables that are (inversely) proportional.
Drawn graphs have labelled and scaled axes, and are used in quantitativemeasurements.
A graph helps in understanding a trend or mathematical relationship in experimentaldata. It shows the effect that changing an independent variable (x-axis) has on thedependent variable (y-axis). The value of the dependent variable depends on the value ofthe independent variable.
Line of best fit
All experimental data is prone to some formof error. A line of best fit shows theappropriate relationship between x and y,taking the variation of individualmeasurements due to error into account.When asked to draw a
line of best fit: use apencil first, andconsider if it shouldpass the origin or not.
The line of best fit does not (necessarily) gothrough all the data points. But it is always acontinuous line without sharp “bends” thatpasses as near to as many points as possible.Lines of best fit can be drawn for anymathematical relationship: linear, quadratic,exponential, etc. . .
Linear graphs
When the dependent variable isproportional to the independent variable, itis called a linear function (i.e. a straightline). The general formula is: y = mx + c
x
y
∆x
∆y
c
gradient =
m
The gradient m (slope) of the line expresses the proportionality factor:→ when m = 2 then y increases by 2 everytime x increases by 1→ when m = 0.43 then y increases by 0.43 everytime x increases by 1
138
MEASUREMENT AND DATA PROCESSING Graphical Techniques 11
The value of the gradient can be calculated by: m =∆y∆x
.
When determining the gradient from a given graph, oftentimes after drawing the line ofbest fit, the triangle used to calculate the gradient should be as large as possible. Also useyour ruler, and take care to read the values from the graph as precisely as you can!
When c = 0, the graph intersects at the origin and y is directly proportional to x.
Graphical techniques
The y-intercept: the initial or starting value, when x = 0.
The gradient of a curve at a certain point�
m =∆y∆x
�
.
Since y/x, the units of the gradient will be the units of y divided by the units of x.The area under the curve between two points (area=∆y ·∆x).
Since y · x, the units of the area will be the units of y times the units of x.Interpolation: a value on a graph between measurements.Extrapolation: a value on a graph outside the measurements, assuming that the graph
continues the trend.
When a graph is not linear, the gradient changes continuously. The gradient at acertain point can be determined by first drawing the tangent, and then the gradientm can be determined from a large∆y vs∆x triangle.
Determine the rate of the reaction at t = 10 s from the given graph, and derivethe units for the rate of reaction.
The rate of reaction is the change in concentration per unit of time, or the gradient inthe concentration vs time graph.
Start by drawing the tangent line at t = 10susing a ruler.Next, find two convenient points on thetangent as far away from each other tocalculate the gradient
m =∆y∆x=
0.30− 0.1025− 0
= 8.0× 10−3
To determine the correct units, simply put theunits into the formula instead of the values:
m =∆y∆x=
moldm−3
s=moldm−3 s−1.
time
[. . . ]
[A]tan
gent
0.1
0.2
0.3
0 10 20 30 40 50
The rate of the reaction at t = 10s is 8.0× 10−3 mol dm−3 s−1.
Exam
ple.
139
MEASUREMENT AND DATA PROCESSING Graphical Techniques
Inversely proportional
A typical question that involves an inversely proportional relationship is that betweenthe pressure and the volume of a gas, when the mass and temperature are kept constant.
From chapter 1, remember that the ideal gas law can be written as: nR=p1V1
T1=
p2V2
T2
So at constant T:p1V1
@@T1=
p2V2
@@T2= p1V1 = p2V2
Since p1V1 = p2V2, when the pressure is doubled: thevolume halves (and vice versa).
This is called inversely proportional, and it produces thegraph you see on the right.
p
V
Such an inversely proportional function can be displayedas a linear function by changing one variable to its inverse.
For example: change the p-axis to1p
.
This technique (as well as manipulation of logs) is alsoused in chapter 5 on the Arrhenius equation, to get a
linear graph between lnA and1T
, with the gradient
m =−Ea
R.
1p
V
140
MEASUREMENT AND DATA PROCESSING Spectroscopic identification 11
11.2 Spectroscopic identification
From personal experience of the authors, we can say that it is often difficult to identifythe compounds that we create in the lab, as well as their purity. Advanced spectroscopictechniques help us identify the compounds we create.
Spectroscopic methods can give insight in the arrangement of atoms that form amolecule. But often there is no certain definitive answer, and we have to understand thateach single technique has strengths and weaknesses.
Oftentimes, chemists combine spectroscopic methods to take advantage of the strengthsand weaknesses of each individual method.
List of spectroscopic methods
The operating principles are not required for any of these methods, you willonly be tested on knowing how and what information to gather.
Index Hydrogen Deficiency (IHD) also known as the degree ofunsaturation, is used to determine the number of H2 moleculesrequired to convert any molecule to one that is saturated and non-cyclic.
Mass spectrometry measures the mass of the molecule, and the masses of thefragments that form from the molecule.
Infrared (IR) spectroscopy identify the type of bonds and functional groupspresent in a molecule.
Proton Nuclear Magnetic Resonance (1H-NMR) spectroscopy identifiesfor hydrogen atoms: 1. their chemical environment, 2. the ratiobetween the number of hydrogen atoms in the same environment, and3. the number of neighbouring hydrogen atoms in a molecule.
X-Ray Diffraction identify the lengths and angles of bonds in any type ofsubstance (metal, ionic and molecular).
141
MEASUREMENT AND DATA PROCESSING Spectroscopic identification
IHD – index of hydrogen deficiency
The value of IDH= number of rings+number of π bonds
It is equal to the number of H2 molecules required to convert any molecule to a saturatedand non-cyclic compound. This means that a saturated and non-cyclic compound wouldhave an IHD value of 0.
The index of hydrogen deficiency is given by:
IHD=2+ 2 ·C+N−H−X
2
C =number of carbon atomsH =number of hydrogen atomsX =number of halogen atomsN =number of nitrogen atoms
When a structural formula of a molecule is given, the IHD is easily determined, namelyby the counting the number of π bonds and rings.
Benzene has an IHD = 4, because it has 3π bonds and one ring.On the exam, students often forget to add one to the IHD foreach ring.
Exam
ple.
When the molecular formula is given the IHD must be determined by the above givenformula. A quick way to remember the formula is by drawing the following structure:
C O C N
+2+2 ·C+N−H−X2
– A saturated non-cyclic molecule starts and ends with ahydrogen, so +2.
– Any carbon atom requires 2 hydrogen atoms each, so+2·C.
– Any nitrogen atom requires 2 hydrogen atoms each, so+N.
– The number of oxygen atoms have no influence on theIHD.
Furthermore, since the IHD is about hydrogen deficiency, we subtract 1 for everyhydrogen we do have. And since halogens replace a hydrogen, we also deduct one forthose.
And finally, since the IHD counts the number of H2 molecules that would be required,we have to divide by 2.
142
